You throw a ball torwards a wall at speed 21 m/s and at an angle 40.0° above the

Question

You throw a ball torwards a wall at speed 21 m/s and at an angle 40.0° above the horizontal (Fig. 4-35). The wall is 22.0 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?
m/s (horizontal)
m/s (vertical) (d) When it hits, has it passed the highest point on its trajectory?

yes

no    

not enough information to decide

Explanation / Answer

b)

here

distance = speed * time

t = 22 / 21 * cos(40deg)

then by using the second equation of motion

y = u*t + 0.5 *a *t^2

y = 21 *cos(40deg) * (22 / 21*cos(40)) + 0.5 * -9.8 * ( 22 / 21*cos(40deg))

y = 14.3 m

c)

vx = 21 * cos(40deg)

vx = 16.08 m/s

this is the horizontal component 16.08 m/s

then the vertical component is

vy^2 = uy^2 + 2 * a* s

vy^2 = (21 * sin(40))^2 + 2 * -9.8 * -14.3

vy = 21.5 m/s

the vertical component is 21.5 m/s

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