The gravitational force of a star on an orbiting planet 1 is F1. Planet 2, which is twice as massive as planet 1 and orbits at t
1 answer:
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Answer:
Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.
Initial speed of the projectile is v and final speed is 0.5 v.
g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :
So, the height of the projectile above the ground is . Hence, this is the required solution.
Answer:
38.4 m/s
Explanation:
a) at t = 3.2s.
b) at t = 3.2 + Δt.
c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt
As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s