Question

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Answer 1

Answer:

22.1 m

Explanation:

$v_{o}$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball  along the vertical direction.

$v_{oy}$ = initial speed of ball = $v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}$

$a_{y}$ = acceleration due to gravity = - 9.8 ms⁻²

$t$  = time of travel

$y$  = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_{ox}$ = initial speed of ball = $v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}$

$X$  = Horizontal distance traveled

$t$  = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m$