how do I solve the following problem?

Question

A cannon on a 12 m wall. Shoots a cannon ball at 37 degrees to a height of 32m above the cannon. What is the initial velocity exiting the cannon? How long does it take to return to the ground 12m below the cannon?

Answer 1

Hi Sherisse

Here is the method 

 

Max height = v²Sin²θ/ 2g 

where v = initial velocity 

given Max height = 32 m

θ = 37 degree

 

32 = v² sin² 37 /2g

gives v= 1.31 m/s

v= sqrt 1.732 m/s 

 

Vertical component of velocity = vcosθ =1.32 Cos 37 = 1.054 m/s

Now for time to hit the ground

Now for total time t₁ = time from 12 m to 32 m , with initial speed of 1.054 m/s

t₂ = time from 32 +12 = 44 m to ground ( initial speed being zero at 44 m from thr ground) 

Total time = t₁ +t₂

t₁= 1.054/9.8 = 0.107 sec

t₂ = sqrt 44*2/9.8  ( using H =1/2 gt²)

=3 sec approx

total time to the ground = t₁ +t₂