Answer:
The ball was hit with the speed of 43.31m/s, and with 52m above the fence
Step-by-step explanation:
Using law of projectile motion,
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
