A homerun ball traveled 188 m before landing outside the ballpark. Assuming the

Question

A homerun ball traveled 188 m before landing outside the ballpark. Assuming the ball's initial velocity was 51 degrees above the horizontal, what did the initial speed of the ball need to be if the ball was hit 0.9 m above the ground? How far would the ball be above a fence 3.0 m high if the fence was 116 m from home plate?

Explanation / Answer

We have to determine the flight time. Ignoring air resistance, the time from the launch height to max height is equal to the time from max height back to launch height. At max height, Vy(t) = 0. Vy slows to a stop before falling at g. Vy(t) = Vyo - gt => Vyo/g = t where t is the time to max height from launch height The time to fall back to the launch height is t = Vyo/g At the moment the ball reaches the launch height Vxf=Vxo and Vyo = -Vyo At h=0.9m the ball has not traveled 188m. We we can draw a right triangle with height 0.9, angle 56 degrees because ignoring air resistance, the impact angle is 56 degrees from the negative x axis. Therefore, the difference in the horizontal distance at the launch height is x and 0.9/x = tan56 so x = 0.9/tan56 = 0.607m 188-0.607=187.4m which is the distance traveled in 2t = 2Vyo/g at Vxo tan56=Vyo/Vxo => Vyo = Vxotan56 Vxo*2t = Vxo*2*Vxo*tan56/g = 187.4 => sqrt(187.4*g/2*tan56) = Vxo=24.9m/s Vyo=36.9 and Vo = 44.5m/s at d = Vxo*t = 116 => t = 116/Vxo = 4.66s h(t) = 0.9 + 44.5*sin56 * 4.66 - 4.9*4.66^2 = 66.4m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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