Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
Answer:
by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2
Answer:
The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.
Explanation:
Answer:
Gravity? is it multiple choice?
