Question

A baseball is hit 40 m/s at an angle of 30 degree. A 5 m high fence is 125 m away. Does it go over the fence?

Answer 1

v_o = initial speed of the baseball at the time launch = 40 m/s

= angle of the launch = 30 deg

Consider the motion of the ball along the X-direction or horizontal direction :

v_ox = initial velocity along the horizontal direction = v_o Cos = (40) Cos30 = 34.6 m/s

a_x = acceleration in horizontal direction = 0 m/s²

X = horizontal displacement = distance of the fence = 125 m

t = time of travel

Using the kinematics equation

X = v_ox t + (0.5) a_x t²

125 = (34.6) t + (0.5) (0) t²

t = 3.6 sec

Consider the motion of the ball along the Y-direction or vertical direction :

v_oy = initial velocity along the vertical direction = v_o Sin = (40) Sin30 = 20 m/s

a_y = acceleration in vertical direction = - 9.8 m/s²

y = vertical displacement = ?

t = time of travel = 3.6 sec

Using the kinematics equation

y = v_ox t + (0.5) a_x t²

y = (20) (3.6) + (0.5) (- 9.8) (3.6)²

y = 8.5 m

h = height of the fence = 5 m

Since y > h

hence the ball goes over the fence.