A home run is hit in such a way that the baseball just clears a wall 12 m high,

Question

A home run is hit in such a way that the baseball just clears a wall 12 m high, located 116 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.) (a) Find the initial speed of the ball. m/s b) Find the time it takes the ball to reach the wall. (c) Find the velocity components of the ball when it reaches the wall. x-component m/ y-component Find the speed of the ball when it reaches the wall. m/s m/s

Explanation / Answer

a)

consider the motion along the X-direction

Vox = initial velocity = vo Cos33

X = horizontal displacement = 116

t = time of travel = ?

using the equation

X = Vox t

116 = (vo Cos33) t

t = 116 /(vo Cos33) eq-1

consider the motion along the Y-direction

Voy = initial velocity = vo Sin33

a = acceleration = - 9.8

Yo = initial Position = 1 m

Yf = final Position = 12 m

using the equation

Y = Yo + Voy t + (0.5) a t2

12 = 1 + (vo Sin33) (116 /(vo Cos33)) + (0.5) (- 9.8) (116 /(vo Cos33))2

vo = 38.2 m/s

b)

using eq-1

t = 116 /(vo Cos33)

t = 116 /((38.2) Cos33)

t = 3.62 sec

c)

Along the X-direction , final velocity as it reach the wall is given as

Vfx = Vox = vo Cos33 = (38.2) Cos33 = 32.03 m/s

Along the Y-direction , final velocity as it reach the wall is given as

Vfy = Voy + at

Vfy = (38.2 Sin33) + (- 9.8) (3.62)

Vfy = - 14.7 m/s

d)

speed at the wall is given as

Vf = sqrt(Vfx2 + Vfy2) = sqrt((32.03)2 + (- 14.7)2) = 35.2 m/s

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