Question

a ball is shot at an angle of 45 degrees into the air with initial velocity of 45 ft/sec. assuming no air resistance, how high does it go? how far away does it land? hint: the acceleration due to gravity is 32 ft per second squared.

Answer 1

Explanation: GIVEN :

∅ = 45°

INITIAL VELOCITY (V·) = 45 FT/SEC

ACC DUE TO GRAVITY = 32 FT/SEC²

REQUIRED :

1. MAXIMUM HEIGHT (HMAX) = ? OR how high does it go 2.MAXIMUM RANGE (RMAX) = ? OR how far away does it land?

BY APPLYING FORMULA OF MAXIMUM HEIGHT OF PROJECTILE

HMAX = V·²SIN²∅/2G

HMAX = 45²SIN45²/2X32

HMAX = 1012.5/64

HMAX = 15.82FT

FOR MAX RANGE

BY APPLYING FORMULA OF MAX RANGE

R = V·²SIN2∅/G

R = 45²SIN2X45/32

R = 2025 X 1/32

R = 63.28FT