Acceleration of block = net force/mass = 10/2 = 5 m/s²
this is a link to a web sight with a diagram to help you
https://www.google.com/url?sa=i&source=images&cd=&ved=2ahUKEwjw4v7knNDgAhVyUN8KHWgWD-wQjRx6BAgBEAU&url=%2Furl%3Fsa%3Di%26source%3Dimages%26cd%3D%26ved%3D%26url%3Dhttps%253A%252F%252Ffuturism.com%252Fwhere-do-all-the-elements-come-from%26psig%3DAOvVaw19_FOCuWs_nMsyY1YT0Da-%26ust%3D1550955269844922&psig=AOvVaw19_FOCuWs_nMsyY1YT0Da-&ust=1550955269844922
Explanation:
First, find the velocity of the projectile needed to reach a height h when fired straight up.
Given:
Δy = h
v = 0
a = -g
Find: v₀
v² = v₀² + 2aΔy
(0)² = v₀² + 2(-g)(h)
v₀ = √(2gh)
Now find the height reached if the projectile is launched at a 45° angle.
Given:
v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)
v = 0
a = -g
Find: Δy
v² = v₀² + 2aΔy
(0)² = √(gh)² + 2(-g)Δy
2gΔy = gh
Δy = h/2
Answer:
a) It reduces the kinetic energy loss of the stunt person.
