Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


"Constant velocity" is practically a definition for zero acceleration.
Answer:
0.5m/s2
Explanation:
acceleration= change in velocity/time taken
= v - u/ t
= 10-5/10
=5/10
= 0.5m/s2
We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.
The stratosphere is the layer above the troposphere