Problem 1: A soccer ball is hit from the ground toward a wall with an initial sp

Question

Problem 1: A soccer ball is hit from the ground toward a wall with an initial speed 25 m/s and at an angle of 40° above the horizontal. The wall is distance d-22 m from the release point of the ball. a) How long does it take for the ball to hit the wall? b) How high from the ground will the ball hit the wall? c) What is the velocity of the ball (magnitude and direction) when it hits the wall? d) What is the maximum height that the ball may reach and how long would it take to reach it? e) When it hits, has it passed the point of maximum height on its trajectory? Explain. Projectile Motion On x-axis: and On y-axis: and

Explanation / Answer

Here,

theta = 40 degree

u = 25 m/s

d = 22 m

a) time taken for ball to hit the wall = d/(u * cos(theta))

time taken for ball to hit the wall = 22/(25 * cos(40 degree))

time taken for ball to hit the wall = 1.15 s

b)

height of the ball = u * sin(theta) * t - 0.50 gt^2

height of the ball = 25 * sin(40 degree) * 1.15 - 0.50 * 9.8 * 1.15^2

height of the ball = 12 m

c)

for the magnitude of velocity of ball

v^2 - 25^2 = -2 * 9.8 * 12

v = 19.7 m/s

the velocity of the ball magnitude is 19.7 m/s

Now, for the angle a

19.7 * cos(a) = 25 * cos(40 degree)

a = 13.6 degree

the angle of the velocity is 13.6 degree below horizontal

d)maximum height = (u * sin(theta))^2/(2g)

maximum height = (25 * sin(40))^2/(2 * 9.8)

maximum height = 13.2 m

time taken to reach there = u * sin(theta)/g

time taken to reach there = 25 * sin(40 degree)/9.8

time taken to reach there = 1.64 s

b) No , as the time taken to reach the maximum height is more than the time taken to collide on wall

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