Question

A rock is launched at a 50-degree angle above the horizontal with

an initial velocity of +16 m/s.
a. Calculate the rock's maximum height.

Answer 1

Step-by-step explanation:

Vertical component of velocity

= 16 sin 50 = 12.3 m/s

Vertical height will be given by

h = vo t + 1/2 a t^2

h = 1/2 ( -9.81) t^2 + 12.3 t

h = - 4.905 t^2 + 12.3 t

will have max at t = - b/2a = -12.3/(2*(-4.905) ) = 1.25 sec

use this value of 't' in the equation to find the max height:

h = - 4.905 ( 1.25^2) + 12.3 ( 1.25) = 7.7 meters max height