Question

a projectile is fired with a velocity of 320 m/s at an angle of 30 to the horizontal. find the time to reach it's greatest height,

Answer 1

Explanation:

Given: angle θ = 30˚ ; Initial velocity = 20m/s

Solve for maximum height dymax = ?

There are three equations to do this. The first equation is solving for the

initial vertical velocity; the second equation is solving for the

time of rise and the third equation is solving for the maximum height.

Solving for the initial vertical velocity :

initial vertical velocity = initial velocity * sin 30

initial vertical velocity = 20 m/s * 0.5

initial vertical velocity = 10 m/s

Solving for the time of rise

time_up = initial vertical velocity / acceleration due to gravity

time_up = 10 m/s / 9.8 m/s^2

time_up = 1.0204 seconds

Solving for the maximum vertical height of rise

dymax = Viy * t – 1/2 gt^2

dymax = 10 m/s * 1.0204 s – 4.9 m/s^2 * (1.0204 s)^2

dymax = 10.204 m – 4.9 * 1.0412 m

dymax = 10.204 m – 5.10196 m

dymax = 5.10 m

The body rises up to a maximum height of 5.10 meters.