Exercises on Projectile Motion

2012-09-17T17:08:24Z

Lentonk: /* Exercise 11 */

''Helena Dedic''

==Exercise 1==

An object is launched from the ground at some angle above horizontal. Three seconds later, its velocity v is (20 m/s, -4 m/s). Find how far (horizontally) and how high the object will fly.

*[[Proj Motion EX 1|SOLUTION Ex 1]] 
 

==Exercise 2==

A balloon is drifting horizontally with a speed of 15 m/s. A bolt drops from the balloon at a height of 100 m above the ground. Use the v - t graphs to solve this problem. 

When will the bolt land on the ground? 

How far from the drop point will it land? Compute the horizontal distance. 

What is the bolt's velocity 2 s into the flight? 

With what velocity will it land? 

*[[Proj Motion EX 2|SOLUTION Ex 2]] 

==Exercise 3==

Amorosa, who is on a balcony 40 m above the ground, throws the key to her heart at 37° below the horizontal to Pedro on the ground. He catches the key two seconds after it was thrown. 

How far from the base of the building was he standing? 
At what angle was the key moving when he caught it? 

*[[Proj Motion EX 3|SOLUTION Ex 3]] 
 

==Exercise 4==

A rocket launched from rest at <math>70^\circ</math> to the horizontal has a constant net acceleration of 8 m/s2 along this direction for 6.5 s and then is in free fall. Find its maximum height and its range. 

*[[Proj Motion EX 4|SOLUTION Ex 4]] 
 

==Exercise 5==

A tennis ball is served at a height of 2.4 m at 30 m/s in the horizontal direction. The net is 0.9 m high at a horizontal distance of 12 m. Does the ball clear the net? If so, by how much? If not, where does it strike the net? 

*[[Proj Motion EX 5|SOLUTION Ex 5]] 
 

==Exercise 6==

The path of a projectile is given below. Sketch the position, velocity and acceleration vectors for the projectile in the position shown and write the components of acceleration. 

[[image:Proj Motion_EX6.png|TOP]] 

*[[Proj Motion EX 6|SOLUTION Ex 6]] 
 

==Exercise 7==

Jack, standing on a cliff, throws a rock into a lake 20 m below with initial velocity 25 m/s at 53° above horizontal. 

How high above the water will the rock rise? 
How far from the cliff (along horizontal) will the rock land? 
With what velocity will it land? 

*[[Proj Motion EX 7|SOLUTION Ex 7]] 
 

==Exercise 8==

A plane flying horizontally dropped a supply of blankets for stranded hunters. The blankets landed 4 s after the drop with a speed of 50 m/s. 

What were the components of the velocity when the supply landed? 
What was the velocity of the plane? 
How high was the plane flying? 
How far did the blankets move in the horizontal direction? 

*[[Proj Motion EX 8|SOLUTION Ex 8]] 
 

==Exercise 9==

A balloon is launched from rest from the ground. By the time it rises to a height of 40 m its velocity is 20 m/s in the direction of 37° above horizontal. Let us choose to call this time t = 0 and this horizontal position x = 0. Assume the balloon maintains this velocity and that, as it reaches 100 m above ground, a bolt drops from the balloon. Use v - t graphs to solve the following: 

When will the bolt hit the ground counting from what we called t = 0? 

How far away will it land? Compute horizontal distance counting relative to x = 0. 

With what velocity will it land? 

*[[Proj Motion EX 9|SOLUTION Ex 9]] 

==Exercise 10==

A ball rolls of a 1 m-high table and lands 1.6 m along the floor. Find its initial speed and the time of flight. 

*[[Proj Motion EX 10|SOLUTION Ex 10]] 

==Exercise 11==

A projectile fired from the ground has a velocity of <math>24\hat i -8\hat j\ m/s </math> at a height of <math>9.8 m</math>. Find the initial velocity and the maximum height. 

*[[Proj Motion EX 11|SOLUTION Ex 11]] 

==Exercise 12==

A ball is thrown from ground level. Three seconds later it is moving horizontally at 15 m/s. Find the horizontal range and the angle of impact. 

*[[Proj Motion EX 12|SOLUTION Ex 12]] 
 

*[[Projectile Motion|Back to Projectile Motion]]

Proj Motion EX 4

2012-09-17T16:59:16Z

Lentonk:

''Helena Dedic''

A rocket launched from rest at <math>70^\circ</math> to the horizontal has a constant net acceleration of 8 m/s2 along this direction for 6.5 s and then is in free fall. Find its maximum height and its range. 

'''Solution:''' 
Note: This solutions uses g=10m/s/s 

A rocket is launched at <math>70^\circ</math> above horizontal presumably from rest on the ground. It has a constant acceleration of 8 m/s/s in this direction. 

It means that the rocket is gaining velocity in both the horizontal and vertical direction. 

We choose the origin at the point on the ground where the rocket is launced. The y-axis is positive upward and the x-axis is positive towards the right in the direction the rocket moves. 

The rocket accelerates for 6.5 s. At this instant both the x-component and y-component of the velocity are positive. 

After that the rocket is in a free fall. It means that the x-component of the velocity is constant and the y-component decreases at a rate of 10 m/s every second. 

When the y-component of the velocity becomes zero the rocket is at its maximum height and the direction of its velocity is horizontal. Then the vertical component of the velocity continues to decrease, the y-component of the velocity becomes negative and the rockets speeds up towards the ground. 

[[image: Proj Motion Sol 4.png|TOP]] 

During the first 6.5 s the rocket gains in both components of the velocity because the acceleration has two non-zero components. 

<math>a_x</math> = 8 cos70° = 2.7 m/s/s 

<math>a_y</math> = 8 sin70° = 7.5 m/s/s 

Now we can determine the velocity at the end of this interval assuming the initial velocity to be zero: 

<math>v_x</math> = 2.7 × 6.5 = 17.6 m/s/s 

<math>v_y</math> = 7.5 × 6.5 = 48.8 m/s/s 

Now we are ready to draw the two graphs. 

The <math>v_x</math> - t graph is initially a straight line with a slope of 2.7 m/s/s and after t = 6.5 s the graph becomes a flat line because the horizontal component of the velocity is constant in free fall. 

The <math>v_y</math> - t graph is initially a straight line with a slope of 7.5 m/s/s and after t = 6.5 s the graph is a straight line with the slope equal to - 10 m/s/s. See the graphs. 

[[image: Proj Motion Sol 4b.png|TOP]] 

The velocity decreases from 48.8 m/s 10 m/s every second after t = 6.5 s. How long does it take before the y-component becomes zero? 

It takes 4.88 s. 

Therefore it takes a total of 6.5 + 4.88 = 11.4 s for the rocket to reach the highest point in its path. 

We can determine the maximum height from the graph. It is the area under the graph ½ × 11.4 × 48.8 = 278 m. 

Now we need to determine the horizontal range. We can find it from the <math>v_x</math> - t graph if we know how long is the rocket in the air. We know that it took 11.38 s to reach the maximum height of 278 m. 

Now we have to find out an interval of time t the rocket moves down: the displacement is equal to - 278 m. The displacement is the area of the triangle and so - 278 = ½ <math>\Delta t (- 10 \Delta t</math>). From this equation we find that <math>\Delta t</math> = 7.5 s. 

Thus the total time in the air is 11.4 + 7.5 = 18.9 s. 

Now the area under the <math>v_x</math> - t graph is equal to ½ × 6.5 × 17.6 + 17.6 × (18.9 - 6.5) = 57.2 + 218.2 = 275.4 = 275 m. 

*[[Projectile Motion|Back to Projectile Motion]] 
*[[Exercises on Projectile Motion|Back to Exercises on Projectile Motion]]

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Exercises on Projectile Motion

Proj Motion EX 4