Question

When Babe Ruth hit a homer over the 7.5-m-high right-field fence 95 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 38 degree angle with the ground.

Answer 1

Vy^2 = Voy^2 + 2ay (y - yo)
-Voy^2 = 2 (-9.8) (6.5 - 1)
Voy^2 = 107.8
Voy = 10.4

Vmin = 10.4 / sin38
Vmin = 16.9 m/s

Answer 2

wait Vy isnt zero hold on

Answer 3

95m /2 = 47.5m

47.5m = Vocos38t

Vy = Voy -9.8t
0 = Vosin38 - 9.8t
0 = Vosin38 - 9.8(47.5 / Vocos38)
0 = Vo^2(0.62) - 590.7
Vo = 30.87 m/s

thats the correct answer

Answer 4

Ill be using this equation twice once for x componet and the nfor y component\[r_f =r_i+v_i t +\frac{1}{2}at^2\]

looking at x component first\[95=0+v \times \cos \theta \times t + \frac{1}{2}(0)t^2\]\[95=v \times \cos \theta \times t\]\[t= \frac{95}{v \times \cos \theta }\]

now y component
\[7.5 = 1 +v \times \sin \theta \times t + \frac{1}{2}(-9.8)t^2\]\[6.5 = v \times \sin \theta \times t -4.9t^2\]now we sub in t from x component\[6.5 = v \times \sin \theta (\frac{95}{v \times \cos \theta}) -4.9 (\frac{95}{v \times \cos \theta})^2\]\[6.5 =95 \tan \theta -4.9 (\frac{95^2}{v^2 \times \cos^2 \theta})\]\[ (95 \tan \theta -6.5)v^2 =4.9 (\frac{95^2}{ \cos^2 \theta})\]\[v =\sqrt{4.9 (\frac{95^2}{ \cos^2 \theta})\times \frac{1}{ (95 \tan \theta -6.5)}}\]sub in 38 degrees for theta
\[v = 32.4283 \frac{m}{s}\]