Question

A cannon fires a projectile at 75 m/s at an angle of 56.7 above the horizontal. How high does the projectile get off the ground?

Answer 1

See <b>Related Questions</b> below.

Answer 2

S=displacement
g= gravitational pull(9.81 m/s²)
*Formula used is derived from here:

--> S= Vf² - Vi²/2a

2as= Vf² - Vi²
Vf²= Vi² + 2as
Change "a" to "g" since trajectory motion ang proble m and "s" ng "h" so magiging
(Vf²= Vi² + 2gh)
Since h=? And initial velocity is Zero
0=(VSinθ)² + 2gh
Divide both sides by 2(-g), since the gravity is pulling downward therfore the sign of constant g will be -9.81m/s²
h= - V² Sin²θ / 2(-g)
h= -(75 m/s)² Sin ² (56.7) / 2(-9.81 m/s²)
h= -3929.478 m²/s² / -19.62 m/s²
*Cancel all the excess units "m" will be left
h= 200.279 meters(ans.)