A ball is thrown straight up with an intial velocity of 85 ft/s. At what times will it have a height of 200 ft?
h=vi*t-1/2 g t^2
200=85*t-1/2 *33*t62 solve for time(s) t. use the quadratic formula.
When 85 t - (g/2) t^2 = 200
g = 32.2 ft/s^2
Do the math. There will be two answers.
To find the times at which the ball will have a height of 200 ft, we need to use the kinematic equations of motion. Let's start by analyzing the motion of the ball.
The equation that relates the displacement, initial velocity, time, and acceleration of an object in free fall is:
s = ut + (1/2)at^2
Where:
s = displacement (height)
u = initial velocity
t = time
a = acceleration (in this case, the acceleration due to gravity, which is approximately -32 ft/s^2)
We want to find the time when the height is 200 ft, so we can plug these values into the equation:
200 = 85t + (1/2)(-32)t^2
Simplifying this equation will give us a quadratic equation that we can solve for t.
(1/2)(-32)t^2 + 85t - 200 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula.
Once we solve for t, we can find the times when the ball will have a height of 200 ft.
Note: Since the ball is being thrown straight up, we can ignore the negative root of the quadratic equation, as it represents the time when the ball is on its way down. We are only interested in the time when the ball reaches a height of 200 ft on its way up.