Question

FIGURE 9 shows a ball being thrown from the top of a building towards a wall 60 m away. The initial velocity of the ball is 20 m s¹ at 40° to the horizontal.How much time does it take to hit the wall? b)What is the distance between P and the position of the ball strike the wall? What is the speed of the ball when it strikes the wall ?

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Answer 1

Vx = 20 m/s * cos 40 = 15.3 m/s horizontal speed

Tx = 60 m / 15.3 m/s = 3.92 s

1) 3.92 s to reach wall

Sy = Vy T - 1/2 g T^2

Vy = 20 * sin 40 = 12.9 m/s initial vertical speed

Sy = 12.9 * 3.92 - 1/2 * 9.8 * 3.92^2 = -24.7

Ball strikes wall at -24.7 m from where it was thrown

Distance between start and end point

D = (60^2 + 24.7^2)^1/2 = 64.9 m

2) Ball strikes wall at 64.9 m from starting point

Final vertical speed Vy2 = Vy - g T

Vy2 = 12.9 - 9.80 * 3.92 = -25.5 m/s

V = (Vx^2 + Vy2^2)^/12 = (15.3^2 + 25.5^2)^1/2 = 29.7 m/s

3) Ball travelling at 29.7 m/s when it hits wall