Question

A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?

Answer 1

a)the ball will leave the bat at an angle of 61.3° .

b) the velocity at which it will hit the ground will be v = 27.1 m/s

Step-by-step explanation:

Given,

v = 47.24 m

h = 0.42 m

t = 5.73 s

R = 130 m

a)We know that

R = v cosθ × t

cosθ =
`(R)/(v t ) = (130)/(47.24* 5.73 ) =0.4803`

θ = 61.3°

the ball will leave the bat at an angle of 61.3° .

b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s

v = u + at

Vy = 47.24 x sin 61.3 - 9.81 x 5.73

= -14.8 m/s

v =
`√(v_x^2 + v_y^2))`

v =
`√(22.7^2 + -14.8^2)`

v = 27.1 m/s

the velocity at which it will hit the ground will be v = 27.1 m/s