An artillery shell is fired from a cliff 20.5 m above a valley with a velocity of 209 m/s at an angle of 30.5 degrees above the horizontal.

Question

How many seconds after being fired does the shell reach its maximum height?

Answer 1

This question is about projectile motion. The most significant fact about projectile motion is that the only force acting on the object is gravitation and the most common assumption is that there is no air resistance.

Lets make a list of relevant variables:

t=? (s)

initial vertical velocity V_iv= 209sin30.5°

At maximum height V_fv is zero

initial horizontal velocity V_ih = 209cos30.5° (we don't need this value for this problem)

a=g=-9.81ms^-2

Using the equation of motion V_f = V_i + (a x t) and rearranging, we have

(V_f-V_i)/a=t

or (0-209sin30.5)/-9.81=t

i.e. 104.5/9.81=10.7s