Question

Oliver hits a baseball at 3 feet above the ground with an initial speed of 135ft/sec at an angle of 20 degrees with the horizontal.
The outfield wall is 380 feet away and is 15 feet high.

At approximately what height will the ball hit the wall?

A) 3 ft
B) the ball lands in front of the wall
C) the ballet goes over the wall
D) 8 ft

Answer 1

Answer:

• Option B) the ball lands in front of the wall

Explanation:

1. Find the vertical and horizontal components of the initial velocity

a) Horizontal component:

      $v_{0,x}=v_0\times cos(20\º)=135ft/s\times cos(20\º)=126.86ft/s$

b) Vertical component:

      $v_{0,y}=v_0\times sin(20\º)=135ft/s\times sin(20\º)=46.17ft/s$

2. Find the time when the ball runs 380 feet horizontally:

          $x=v_{0,x}\times t\\\\380ft=126.86ft/s\times t$

          $t= (380ft)/(126.86ft/s)=2.995s\approx 3.0s$

3. Find the height of the ball when the time is 3.0s

       $y=y_0+v_{0,y}\times t-gt^2/2$

• y₀ = 3 ft
• g ≈ 32.174 ft/s²

     $y=3ft+46.17ft/s\times 3.0s-32.174ft/s^2\times (3.0s)^2/2$

     $y=-3.3ft$

This result means that the ball does not reach the wall because it has fallen to the ground (when y = 0) before 3.0s.