An object is thrown straight upward from the edge of a building with a velocity of 20 ms-1. Where will the object be 5 s after it was thrown?
22.5 m above the point from which it was thrown
h = Hi + Vi t - 4.9 t^2
h = Hi + 20 * 5 - 4.9 * 25
h = Hi + 100 - 122.5
= Top of building - 22.5 meters
So below, not above
To solve this problem, we need to consider the motion of the object and how it changes with time due to the force of gravity.
We know that the initial velocity of the object when it is thrown upward is 20 m/s. We also know that the acceleration due to gravity is approximately 9.8 m/s^2, directed downward.
Since the object is thrown upward, we can assume that the initial velocity is positive (upward) and the acceleration due to gravity is negative (downward).
To find the object's position after 5 seconds, we can use the formula for the position of an object in free fall:
s = ut + (1/2)at^2
Where:
s is the object's position
u is the initial velocity
t is the time
a is the acceleration due to gravity
Plugging in the values we know:
s = (20 m/s)(5 s) + (1/2)(-9.8 m/s^2)(5 s)^2
s = 100 m - 122.5 m
s = -22.5 m
Therefore, the object will be at a position of -22.5 meters after 5 seconds. The negative sign indicates that the object is below the starting point, which is expected since it is thrown upward and then falls back down due to gravity.