1) A projectile is shot horizontally at 65.3 m/s from the roof of a building 24.

Question

1) A projectile is shot horizontally at 65.3 m/s from the roof of a building 24.4 m tall.

(a) Determine the time necessary for the projectile to reach the ground below.

(b) Determine the distance from the base of the building that the projectile lands.

(c) Determine the horizontal and vertical components of the velocity just before the projectile reaches

the ground.

Explanation / Answer

s= 24.4 => v=u+at but u=0 for vertical direction s=ut+.5gt^2 => 24.4/(.5g)= .50 seconds =t the distance from the base of the building that the projectile lands = v*t as accel in (x)=0 d= v*t=> 65.3*.5 = 130.6m harizontal velocity = 65.3 vertical velocity v= u +gt = 5m/s

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