Question

A home run is hit in such a way that the baseball just clears a wall 16 m high, located 122 m from home plate. The ball is hit at an angle of 33° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.)

What is the initial speed of the ball?

How long does it take the ball to reach the wall?

What is the velocity of the ball (x and y-component)?

What is the speed of the ball when it reaches the wall?

Answer 1

Sol::
Given
Angle= 33°
Wall height h= 16 m
Distance = 122 m

Since the air resistance is negligible,let's calculate the x and y component..

The ball is hit 35° to the horizontal so the relation between the x and y components of the initial velocity is:
tan 33° = vy / vx
vy = vx * tan 33°

vx is constant, but vy changes with time.

Δx = 122m
Δy = 15m (16m - 1m)

Therefore:
Δx = vx*t
Δy = vy*t +1/2*g*t²

Plug in known values:
122m = vx*t
vx = 122m / t

Substitute in Vy then Vx:
15 = vx * tan 33° * t + 0.5* (-9.8) * t²
15 = (122 / t) * tan 33° * t -0.5 * 9.8* t²
15 - 122*tan 33° = -4.9* t²
-4.9*t^2=-64.22
t=√(64.22/4.9)
t = 3.62 s

The ball travels 3.62 s before going over the fence

a)
Now the initial velocity will be

cos 33° = Vx / velocity

Vx = dx / t
= 122/ 3.62
= 33.7 m/s

velocity = Vx / cos 33°
= 33.7 / cos 33°
= 40.18 m/s

b) the time ball takes to reach the wall is
t = 3.62 s

c)
Vx is constant at 33.7 m/s

vy can be calculated using the equation:
v = v0 + a*t

Initial Vy = 40.18 * sin 33°
= 21.88 m/s
Final Vy = 21.88 + (-9.81) * 3.62 s
Final Vy = -13.596 m/s
Vy= -13.6 m/s

The speed when the ball reaches the wall is (using Pythagorean Theorem):
v = √(vx² + vy²) = √((33.7)² + (-13.6 )^2)
v = 36.34 m/s

Answers:
a) Initial ball speed: 40.18 m/s
b) Time for ball to reach wall: 3.62 s
c) x velocity-component: 33.7 m/s
y velocity-component: -13.6 m/s
Speed of the ball: 36.34 m/s

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