How high is the highest point of the trajectory?

Question

A cannon sends a projectile towards a target a distance 1440 m away. The initial velocity makes an angle 23◦ with the horizontal. The target is hit. The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the initial velocity? Answer in units of m/s.
011 (part 2 of 3) 2.0 points How high is the highest point of the trajectory? Answer in units of m.
012 (part 3 of 3) 2.0 points How long does it take for the projectile to reach the target? (Assume no friction) Answer in units of s

Answer 1

You are given the range R and the launch angle θ.  From the kinematics of projectile motion,

 

R = v₀²sin(2θ)/g ⇒

 

v₀ = √[Rg/sin(2θ)] = √[1440(9.8)/sin(46°)] m/s = 140.1 m/s

 

h_max = v₀²sin²θ/(2g) = (140.1)²sin²(23º) / [2(9.8)] m = 152.9 m