Proj Motion EX 12

Helena Dedic

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.

Solution:

At t = 2 s, the velocity of a projectile is v = (15 m/s, 0 m/s). What is the range of the flight and the angle of impact?

Since $v_y=0$, the projectile must be at the top of its motion at t = 2 s. The total time of the motion is then 4 s and the range is given by equation 1:

$\mathrm{\Delta }x=v_{xo}t$ 

$\mathrm{\Delta }x=15m/s*4s=60m$ 

Find the velocity in the y-direction at impact taking the initial velocity as zero at the top and finding the velocity 2 s later:

$v_y=v_{yo}-9.8m/s^{2}t$ 

$v_y=0-9.8m/s^{2}2s$ 

$v_y=-19.6m/s$ 

The velocity at impact is v = (15 m/s, -19.6 m/s) so that the angle of impact is $52.6^o$ below the horizontal or $-52.6^o$.

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