Question

You fire a projectile 35° above horizontal with an initial velocity of 200m/s. It lands in a valley 300m below the launch point. What is the time of flight of the projectile, and what is the range of projectile? ​

Answer 1

See below

Step-by-step explanation:

Vertical component of initial velocity = 200 sin 35° = 114.72 m/s

then use position formula a = 9.81 m/s^2

0 = 300 + 114.72 t - 1/2 (9.81)(t^2)

use quadratic formula with a = - 4.905 b = 114.72 c = 300

to find t = 25.76 seconds

To find the range

( horizontal distance the projectile lands from launch point)

Horizontal component of initial velocity 200 cos 35 = 163.83 m/s

( it flies horizontally at this speed for 25.76 seconds <==found above)

163.83 m/s * 25.76 s = 4220.3 meters