Question

A projectile is launched with an initial velocity vo at an angle theta above the horizontal.
In terms of vo, theta and acceleration due to gravity g, determine for the projectile
i) the time to reach its maximum height and
ii) its maximum height.

Answer 1

We take the upward direction to be positive, and downward direction to be negative.

(i)

Initial vertical velocity is

Vertical acceleration is

We take vertical acceleration as negative because it is directed in the downward direction.

At the maximum height, the vertical velocity of the projectile changes its direction from up to down. At the moment vertical velocity changes its direction from up to down, the vertical velocity is momentarily zero. So, at the maximum height

The vertical acceleration is constant. The vertical motion is governed by the kinematics equation for uniformly accelerated motion. To find time for changing velocity from v_iy to v_y=0 we use

(ii)

Maximum height is equal to the vertical distance traveled by the projectile in time t. Since vertical motion is uniformly accelerated motion, the motion is governed by the kinematics equation for uniformly accelerated motion. To find vertical distance traveled, we use