If a ball dropped from the top of a 400 ft building to hit the ground, how long would it take in seconds?
Express 0.000093 in scientific notation. Can anyone help?
since the height
h(t) = 400-16t^2
h=0 when t=5
To determine the time it takes for the ball to drop from the top of the building to hit the ground, we can use the formula for free fall:
d = 1/2 * g * t^2
Where:
d = distance (in this case, the height of the building) = 400 ft
g = acceleration due to gravity = 32.2 ft/s^2 (approximately)
t = time
Plugging in the values we have:
400 = 1/2 * 32.2 * t^2
Now, let's solve for t.
First, we can multiply both sides of the equation by 2:
800 = 32.2 * t^2
Then divide both sides of the equation by 32.2:
t^2 = 800 / 32.2
Simplifying further:
t^2 ≈ 24.845
Taking the square root of both sides to solve for t:
t ≈ √24.845
Using a calculator, we find that t ≈ 4.984 seconds.
Therefore, it would take approximately 4.984 seconds for the ball to hit the ground.
To calculate the time it takes for the ball to hit the ground, we can use the equations of motion and the value of acceleration due to gravity.
The relevant equation for the vertical motion of the ball is given by:
s = ut + (1/2)gt^2
Where:
- s is the displacement (in this case, the total distance fallen)
- u is the initial velocity (which is zero since the ball is dropped)
- g is the acceleration due to gravity (approximately 32.2 ft/s^2)
- t is the time taken
In this scenario, the ball is dropped from rest, so the initial velocity (u) is zero. The displacement (s) is the height of the building, which is 400 ft. Plugging these values into the equation, we have:
400 = 0 + (1/2) * 32.2 * t^2
To solve for t, we need to rearrange the equation:
400 = 16.1 * t^2
Divide both sides of the equation by 16.1:
t^2 = 400 / 16.1
t^2 ≈ 24.84
Taking the square root of both sides:
t ≈ √24.84
t ≈ 4.98
Therefore, it would take approximately 4.98 seconds for the ball to hit the ground.