Answer:
Explanation:
If v be the velocity just after the rebound
Kinetic energy will be converted into potential energy
1/2 m v² = mgh
v² = 2gh
v = √ 2gh
= √ 2 x 9.8 x .96
= 4.33 m / s
Answer:
y = x tan θ - (g / 2v₀² cos² θ) x²
Explanation:
An equation is called a general formula that relates the position on the x-axis and the height on the body's axis.
Let's write the position on each axis
X axis. No acceleration
x = v₀ₓ t
Y Axis. There is the acceleration of gravity
y =
t - ½ g t²
Let's clear the time in the first equation and substitute in the second
y = v₀ sin θ (x / v₀ₓ) - ½ g (x / v₀ₓ)²
y = v₀ sin θ / v₀ cos θ x - ½ g x² / v₀² cos² θ
y = x tan θ - (g / 2v₀² cos² θ) x²
This is the trajectory equation in projectile launching
I think its friction and gravity change the motion hope this helps :)