Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
Explanation:
We need only to apply the definition of acceleration, which is:
In our case the final velocity is 
, the initial velocity is 
since it departs from rest, the final time is 
and the initial time we are considering is 
So for our values we have:
The answer is c. the force of his swing
At the time of the impact, there is a collision between two bodies moving in opposite directions.
The force exerted on the ball causes the change of velocity.
I think its d. but im not sure
