Answer:
a) 1.75s b) 17.2 m/s (down)
Explanation:
d1= 15m d2= 0m (because it hits ground)
a= -9.81 m/s^2 t=???
Equation
the triangle means change in so d2-d1
Δd= v1 * t + 1/2 * a * t^2
0m-15m= v1*t + 1/2 a t^2
-15 m= 0m/s*t (goes away) + 1/2* a *t^2
-15mx2= t^2
-15mx2/a= t^2
Square root (-30/-9.81m/s^2)
t=1.75 s
b) now v2!!
Im going to use v2= v1 + a*t
v2= 0m/s + -9.81 x 1.75s
v2 = -17.2 m/s or you can say 17.2 m/s down!!!
The impulse given to the ball is equal to the change in its momentum:
J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s
This is also equal to the product of the average force and the time interval ∆t :
J = F(ave) ∆t
so that if F(ave) = 200 N, then
∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s
Answer:

Explanation:
We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

Similarly,

Now, the angle between velocity and acceleration vectors can be found.
The angle between any two vectors can be found by scalar product of them:

So,

At time t = 0, this equation becomes

Answer: Final speed
Explaination: because its final.