Answer:
Explanation:
At the top , horizontal displacement = 2 m .
vertical displacement = 1.5 m .
Let time to reach the hole be t . Let the ball is launched at angle θ with horizontal with velocity u .
ucosθ x t = 2
from v = u - 2gt
0 = usinθ - 2gt
usinθ = 2gt
v² = u² - 2gh
u²sin²θ = 2 g h
u²sin²θ = 2 x 9.8 x 1.5
usinθ = 5.42
usinθ = 2gt
2gt = 5.42
t = .2765 s
ucosθ x t = 2
ucosθ x .2765 = 2
ucosθ = 7.23
usinθ = 5.42
dividing ,
Tanθ = .75
θ = 37°
usin37 = 5.42
u = 9.03 m /s