Answer:
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
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 = [tex]\dfrac{R}{v t } = \dfrac{130}{47.24\times 5.73 } =0.4803[/tex]
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 = [tex]\sqrt{v_x^2 + v_y^2)}[/tex]
v = [tex]\sqrt{22.7^2 + -14.8^2}[/tex]
v = 27.1 m/s
the velocity at which it will hit the ground will be v = 27.1 m/s
