Answer:
(a) Height is 4.47 m
(b) Height is 4.37 m
Solution:
As per the question:
Initial velocity of teh ball, 
Angle made by the ramp, 
Distance traveled by the ball on the ramp, d = 5.00 m
Now,
(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

where
H =
g = 

= 19.06 m/s
Now, maximum height attained is given by:


Height from the ground = 
(b) now, considering the coefficient of friction bhetween ramp and the ball,
:
velocity can be given by the eqn-3 of motion:


= 18.7 m/s
Now, maximum height attained is given by:


Height from the ground = 
Answer:
5.09 m/s
Explanation:
Use the height to find the time it takes to land:
y = y₀ + v₀ᵧ t + ½ gt²
0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.28 s
Now use the horizontal distance to find the initial velocity.
x = x₀ + v₀ₓ t + ½ at²
6.5 m = 0 m + v₀ (1.28 s) + ½ (0 m/s²) (1.28 s)²
v₀ = 5.09 m/s
You get circular motion, where the acceleration is pointing towards the center of the circle, as long as they are constant, and not fluctuating.
Energy (in Physics) is the ability to do work.