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 .
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
To find the height of the projectile from the ground after 2.0 seconds, we need to calculate the vertical distance traveled by the projectile.
Here are the steps to solve the problem:
1. Determine the initial vertical velocity (Vy0) of the projectile. Given that the projectile is shot horizontally, the initial vertical velocity is zero since there is no vertical component to the motion initially.
2. Determine the time of flight (t) of the projectile. The time of flight is the total time the projectile is in the air. Since the projectile is shot horizontally, the time of flight is the same as the time it takes for the projectile to hit the ground. In this case, the time of flight is 2.0 seconds.
3. Use the vertical motion equation: h = Vy0 * t + (1/2) * g * t^2, where h is the height, Vy0 is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Since the initial vertical velocity is zero (Vy0 = 0), the equation simplifies to:
h = (1/2) * g * t^2
4. Substitute the known values into the equation. The acceleration due to gravity on Earth is approximately 9.8 m/s^2.
h = (1/2) * 9.8 m/s^2 * (2.0 s)^2
5. Calculate the height:
h = (1/2) * 9.8 m/s^2 * 4.0 s^2
h = 19.6 m/s^2 * 4.0 s^2
h = 78.4 m
Therefore, the height of the projectile from the ground after 2.0 seconds is 78.4 meters.