Question

A batter hits a pitched ball when the center of the ball is 1.21 m above the ground.The ball leaves the bat at an angle of 45° with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 103 m. (a) Does the ball clear a 8.49-m-high fence that is 93.0 m horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

Answer 1

Solution :

Given :

h = 1.21 m

θ = 45^o

R = 103 m

.

Since the range of the projectile is given by :

∴ g R = u² sin(2θ)

∴ (9.81 m/s²)(103 m) = u² sin(2 x 45)

∴ u² = 1010.13 m²/s²

∴ u = 31.787 m/s

.

So, u_x = u sinθ = (31.787 m/s) sin(45) = 22.48 m/s

u_y = u cosθ = (31.787 m/s) cos(45) = 22.48 m/s

.

Here time taken by the ball to travel 93 m horizontally will be :

t = x / u_x = (93 m) / (22.48 m/s) = 4.14 sec

.

And, Vertical height of the ball after that time will be : h = h₀ + u_y t + (1/2) a_y t²

∴ y = (1.21 m) + (22.48 m/s)(4.14 s) + (1/2)(- 9.81 m/s²)(4.14 s)²

∴ y = 10.207 m

.

Therefore, The ball will clear a 8.49-m-high fence that is 93.0 m horizontally from the launch point.

.

the distance between the fence top and the ball center will be = 10.207 m - 8.49 m = 1.72 m