An object falls from the top of the school, which is 6.06m high. Assume air resistance can be ignored and the acceleration is 9.8m/s^2
If the object was dropped( not thrown) , what is its speed after 0.333s
How long does it take to hit the ground
v = at = -9.8(.333) = -3.26
s = 6.06 - 4.9t^2
when it hits, s=0, so
0 = 6.06 - 4.9t^2
t = 1.11
The speed at t = 0.333 s is
V = (g/2) t^2
It hits the ground when (g/2) t^2 = 6.06 m
Solve for that new t.
To find the speed of the object after 0.333 seconds, we can use the formula:
v = u + at
Where:
v is the final velocity (speed) of the object
u is the initial velocity of the object (0 m/s in this case because it was dropped, not thrown)
a is the acceleration due to gravity (9.8 m/s^2)
t is the time (0.333 seconds in this case)
Plugging in the values:
v = 0 + (9.8 * 0.333)
v = 3.2674 m/s
Therefore, the speed of the object after 0.333 seconds is approximately 3.2674 m/s.
To find the time it takes for the object to hit the ground, we can use the formula:
s = ut + (1/2)at^2
Where:
s is the distance traveled by the object (6.06 m - height of the school)
u is the initial velocity of the object (0 m/s because it was dropped, not thrown)
a is the acceleration due to gravity (9.8 m/s^2)
t is the time we are trying to find
Plugging in the values:
6.06 = 0 * t + (1/2) * 9.8 * t^2
Rearranging the equation:
4.9t^2 = 6.06
Dividing both sides by 4.9:
t^2 = 6.06 / 4.9
t^2 = 1.235
Taking the square root of both sides to find t:
t = √1.235
t ≈ 1.11 s
Therefore, it takes approximately 1.11 seconds for the object to hit the ground.