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riadik2000 [5.3K]
3 years ago
8

A person hits a 45-g golf ball. The ball comes down on a tree root and bounces

Physics
1 answer:
Verizon [17]3 years ago
5 0

Answer:

Maximum height, h = 10 m          

Explanation:

It is given that,

Mass of golf ball, m = 45 g = 0.045 kg

The ball comes down on a tree root and bounces  straight up with an initial speed of 14.0 m/s.

We need to find the height the ball will rise  after the bounce. It is based on the conservation of energy such that,

\dfrac{1}{2}mv^2=mgh

h is maximum height raised by the ball

h=\dfrac{v^2}{2g}\\\\h=\dfrac{(14)^2}{2\times 9.8}\\\\h=10\ m

So, the ball will raised to a height of 10 meters.

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3 0
3 years ago
A spring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45
Naily [24]
When firing straight up:
v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained.

Then,
0 = u^2 - 2gh
u = Sqrt (2gh) ---- (1)

When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45

Maximum height, H = U^2*(Sin Ф)^2/2g

substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
H = [h*(Sin 45)^4] = h/4

Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.
6 0
3 years ago
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