If a ball is dropped from from 288 ft above the ground, how long does it take to reach ground level?
The distance S for a free falling object from rest over t seconds is:
S=(1/2)gt²
where
g=9.8 m.s-2
S=288m
Solve for t.
solve for t.
Be cautious about units on this problem. The problem gives the distance in feet. Gravitational acceleration in English units is about 32 ft/sec^2 .
Oops. Use g=32.2 f s-2
Thanks Qudditch.
To find out how long it takes for the ball to reach ground level, we can use the equation of motion for a falling object. The equation is:
s = ut + (1/2)gt^2
In this equation:
- s represents the distance traveled by the object
- u represents the initial velocity (which is zero if the ball is dropped)
- t represents the time taken by the object
- g represents the acceleration due to gravity (which is approximately 32 ft/s^2)
In this case, since the ball is dropped, the initial velocity (u) is zero. The distance (s) is given as 288 ft. We need to find the time (t). So, rearranging the equation, we get:
288 = 0*t + (1/2)*32*t^2
Simplifying this equation gives us:
288 = 16t^2
Dividing by 16:
18 = t^2
Taking the square root of both sides:
t = √18
Calculating the square root of 18 gives us:
t ≈ 4.2426
Therefore, it takes approximately 4.2426 seconds for the ball to reach ground level when dropped from a height of 288 ft.