The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Answer:
d) What is the force if we doubled both the masses AND we doubled the distance
Answer:
Gravity changes with altitude. as we know The gravitational force is proportional to 1/R2, where R is your distance from the center of the Earth.
eg. The radius of the Earth at the equator is 6400 kilometers.
Let's say you were in a jet at the equator that was 40 kilometers high above the earth's surface.
may be helpfull
Answer:
0.80 m
Explanation:
elastic potential energy formula
elastic potential energy = 0.5 × spring constant × (extension) 2