You throw a ball at an angle of 30 above the horizontal at a wall 16 m away. The

Question

You throw a ball at an angle of 30 above the horizontal at a wall 16 m away. The ball's initial speed is 14 m/s and it leaves your hand at a height of 1.5 m above the ground.

A)How long does the ball take to get to the wall?

B)How high up on the wall does the ball hit?

C)What is the horizontal component of the ball's velocity as it hits the wall? Assume that the direction from you toward the wall is the positive direction for the horizontal component of the velocity.

D)What is the vertical component of the ball's velocity as it hits the wall? Assume that the positive direction for the vertical component of the velocity is upward.

Explanation / Answer

a. the ball take time to get to the wall

vcos = x/t
t = x/(vcos)
t = 16/(14cos30°)
t = 1.319 seconds

b. y(t) = (1/2)at² + vtsin + y0
y(1. 319) = (1/2)(-9.81)(1. 319)² + 14(1. 319)sin30° + 1.5
y(1. 319) = -8.346 + 9.233 + 1.5
y(1. 319) = 2.387 m

c) horizontal component

v_x = vcos
v_x = 14cos30°
v_x = 12.124 m/s
d) vertical component

v_y = at + vsin
v_y = (-9.81)(1.319) + 14sin30°

v_y = -5.939 m/s




Since the vertical component of the velocity (v_y) is negative, the ball is moving down.

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