According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor league game. The ball traveled 188 m (618 ft) before landing on the ground outside the ballpark. Assuming the ball's initial velocity was 51 degree above the horizontal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? Assume that the ground was perfectly flat. Express your answer using two significant figures. How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m (380 ft) from home plate? Express your answer using two significant figures.

Let V be the initial speed of the ball. The ball will move at a constant speed horizonally but accelerates vertically in x-direction, and in the y-direction at angle 51°

Break down the intial speed

Vx = Vcos51

Vy = Vsin51

find the time it takes the ball to hit the ground 188m away horizonally

x = vt

188 = Vcos51 × t

t = 188 / Vcos51

Using the equation

Xf = 1/2at^2 + Vt + Xi

=Xf = .5at^2 + Vt + Xi

Where

Xf = final position (0m)

Xi = initial position (.9m)

a = acceleration (taken as 9.8m/s )

t = time

V = speed in y-derection

Plug in 188 / Vcos51 for time

0 = 0.5(-9.8)(188 / Vcos51)^2 + Vsin51 × (188 / Vcos51) + 0.9

-0.9 = -4.9(188 / Vcos51)^2 + 232.2

-233.1 = -4.9(188 / Vcos51)^2

47.57 = (188 / Vcos51)^2

Square root both sides

6.897 = 188 / Vcos51

6.897 × Vcos51 = 188

V = 188 / (6.897cos51)

V = 43.31 m/s

The ball was hit with the speed of 43.31m/s

B)

Find the it takes the ball to travel 116m using

x = vt

116 = 43.31cos51 × t

t = 4.26s

Xf = 0.5at^2 + Vt + Xi

Xf = 0.5(-9.8)(4.26s)^2 + 43.31 sin51 (4.26s) + .9

= -88.92324 + 144.2838961764

Xf = 55.36m aprx 55m

the ball was 55m - 3m = 52m above the fence