Answer1)https://d2vlcm61l7u1fs.cloudfront.net/media%2Fd57%2Fd572025e-d18f-44e8-bad3-77452bbe8de9%2FphpEhl3uR.png

Answer2)a) Time to reach the maximum height for a projectile is given as,

tmax = u.sin(θ)/g = 9.6*sin(45)/9.8 = 0.69 s

b) Displacement in the Y-direction for a projectile can be found out by using the kinematic equation,

s = ut +1/2at2

∴ Δy = u.sin(θ)tf + 1/2(-g)tf2

∴ -6.5 = 9.6*sin(45)*tf - 0.5*9.8*tf2

So, we get a quadratic equation in tf as,

4.9tf2 - 6.79tf - 6.5 = 0

Solving this for tf we get,

tf = (6.79 +/- √(6.792 + 4*4.9*6.5))/(2*4.9)

∴ tf = 2.04 s or tf = -0.65 s

We can discard the negative solution.

∴ Time taken to hit the ground is 2.04 s

c) Range of the projectile will be nothing but the X-displacement in 2.04 seconds.

Δx = u.cos(θ)*tf = 9.6*cos(45)*2.04 = 13.85 m

Answer3)https://media.cheggcdn.com/media%2F2d9%2F2d96bf5f-5197-4f9a-b80f-5110f5b3a79a%2FphpmIyEOU.png

Answer4)q2 https://d2vlcm61l7u1fs.cloudfront.net/media%2Fe1e%2Fe1edf873-6012-4324-aaef-72971d1a8c3d%2FphpzUxnQS.png

https://d2vlcm61l7u1fs.cloudfront.net/media%2Fae0%2Fae002a30-e6db-489b-832f-f721c33384ea%2FphpQP4DdD.png

q3
https://d2vlcm61l7u1fs.cloudfront.net/media%2F44c%2F44c91215-21cc-4748-8372-ed6565141d62%2FphpelRp8K.png