A home run is hit in such a way that the baseball just clears a wall 16.0 m high

Question

A home run is hit in such a way that the baseball just clears a wall 16.0 m high, located 126 m from home plate. The ball is hit at an angle of 33.0° 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.
s

(c) Find the velocity components of the ball when it reaches the wall.

Find the speed of the ball when it reaches the wall.
m/s

Explanation / Answer

along horizantal

vox = vo*cos33

acceleration = ax = 0

distance travelled = x = 126 m

x = vox*T

T = x/vox   >........(1)

along vertical

voy = vo*sin33

acceleration = ay = -g = -9.8 m/s^2

displacement y = h-1 = 16-1 = 15 m

from the equations of motion

y = voy*T + 0.5*g*T^2 ........(2)

from 1 & 2

y = voy*x/vox + 0.5*g*x^2/vox^2

15 = tan33*126 - 0.5*9.8*126^2/(vox^2)

vox = 34.12 m/s

vo*cos33 = 34.12

vo = 34.12/cos33 = 40.7 m/s     <<------------answer

++++++++++++++++++++++

(b)

time t = x/vox = 126/34.12

t = 3.7 s   <<---------answer

+++++++++++++++++

(c)

Vx = vox + ax*T = vox = 34.12 m/s

vy = voy + ay*T

vy = (40.7*sin33)-(9.8*3.7)

vy = -14.1 m/s

speed = v = sqrt(vx^2+vy^2)

v = 36.9 m/s <<-----------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.