Question

an object is launched from ground level with an initial velocity of 400 ft per second. to the nearest tenth of a second, how long does it take the object to reach the height of 650 ft?

Answer 1

\[d=V0t+\frac{1}{2}at^2\]where d = distance, a = acceleration, V0 = initial velocity and t = time.

The only acceleration in the y direction is the acceleration due to gravity (-32ft/s^2).

To solve the problem just fill in the values and solve for t:
\[650ft=(400ft/s)t+12(−32ft/s^2)t^22\]
This can be seen more clearly by removing the units and simplifying it a bit:
\[650=400t−16t^2\]
Now just solve for t...you'll get two values:
- The lower value will be the time it takes to reach 650ft on the way up.
- The higher value will be the time it takes to fall back down to 650ft.