You throw a ball towards a wall at speed 23 m/s and at an angle 40.0° above the

Question

You throw a ball towards a wall at speed 23 m/s and at an angle 40.0° above the horizontal. The wall is 21.2 m from the release point of the ball.

(a) How long does the ball take to reach the wall?
s

(b) How far above the release point does the ball hit the wall?
m

(c) What are the horizontal and vertical components of its velocity as it hits the wall?

(d) When it hits, has it passed the highest point on its trajectory?

YesNo     not enough information to decide

Explanation / Answer

Solution:

Horizontal distance = horizontal velocity * time

tie taken = t = Vo cos 40 / d = 23 cos 40 / (21.2)

= 0.83 s

b) to find the height itreaches , kinematic equation can be used.

y = voy t -1/2gt^2 = 23 sin 40 * 0.83 - 1/2 * 9.8 * 0.83^2= 8.89 m

c) WHen it hitthe wall, its horizontal velocity remains as vo cos40 since gravity does not influence horizontal motion.

vx = vo cos 40 = 23 cos 40 = 17.6 m/s

vertical velocity = vf y = voy - gt =23 sin40 - 9.8 * 0.83 = 6.65 m/s

d) Highest point of its trajectory is when the final velocity in y direction becomes 0 .

vfy^2 = voy^2 -2gH

H = (23 sin 40)^2 /2*9.8) = 11.15 m= highest point of its trajectory

To hit the wall, it need not pass thru the highest point of its trajectory. Wall is at a lesser height than the highest point.

NO

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