Need help on a physics problem

Question

A cannon sends a projectile towards a target a distance 1360 m away. The initial velocity makes an angle 34° with the horizontal. The target is hit. The acceleration of gravity is 9.8 m/s^2 What is the magi itude of the initial velocity? How high is the highest point of trajectory? How long does it take for the projector to reach the target? (assume no friction)

Answer 1

This is a complex problem that does not seem to have enough information to solve at first. However, with multiple steps we can get the answers.

First, we can write equations for the horizontal (vx) and vertical (vy) components of the initial velocity (v): cos 34 = vx/v; sin 34 = vy/v and vx = 0.829v; vy = 0.559v

Next we can analyze the time it takes to reach the highest point and this is the time it takes to travel halfway to the target horizontally: vy = 9.8t; vxt = 1360/2 = 680 m, so vx = 680/t

Substitute trig functions into these last two: 0.559v = 9.8t ; 0.829vt = 680

Now we have two equations and two unknowns, solve for v and substitute: v = 17.53t; (0.829)(17.53t)t = 680; t² = 46.79; t = 6.84 s; then find v = (17.53)(6.84) = 119.9 m/s

This is half the time it takes to reach the target so that time is 13.7 s.

We can use the time to the top and h = 1/2gt² to find the maximum height: h = 4.9(6.84)² = 229.3 m