Question
A projectile is shot horizontally at 34.5m/s from the edge of a 55.0metre building . Determine the height of the projectile from the ground after 2.0seconds .
Answers
To calculate the height from the ground, you only need to consider the vertical component of motion, not the horizontal one.
Since the initial velocity is horizontal, the initial vertical velocity is zero.
u = 0 m/s, t = 2 s, a = -9.8 m/s^2
Use the following equation of motion:
s = ut + (1/2)at^2
=> s = 0 + (1/2)(-9.8)(2^2)
=> s = -19.6 m
Hence, it moves 19.6 m downwards.
Height from the ground = 55 - 19.6
= 35.4 m
Since the initial velocity is horizontal, the initial vertical velocity is zero.
u = 0 m/s, t = 2 s, a = -9.8 m/s^2
Use the following equation of motion:
s = ut + (1/2)at^2
=> s = 0 + (1/2)(-9.8)(2^2)
=> s = -19.6 m
Hence, it moves 19.6 m downwards.
Height from the ground = 55 - 19.6
= 35.4 m