A bullet was fired with a 10 m/s horizontal velocity and a 29.4 m/s vertical velocity. The total time of the bullet's flight was 6 secs.

Question

What was the maximum height and maximum range of the bullet? I got the answers of Δy=44.1 m And Δx= 60.0 m

Answer 1

Assuming it was launched from the ground and hit the ground at t = 6 sec,

 

Δx = v_0xt= (10.0 m/s)(6 s) = 60.0 m

 

Because of the symmetry of this problem, launching from and returning to the same height, the maximum height is reached at t = half of the total time of flight.

 

h(t) = -4.9t² + 29.4t

 

t = 6/2 s = 3 s

 

h_max = h(3) = [-4.9(3²) +29.4(3)] m = 44.1 m  

 

You are correct.