Question

The height of a ball thrown directly up with a velocity of 10 feet per second from a initial height of of 100 feet is given by the equation h(t) = -16tsquared + 10t + 100, where t is the time in seconds and h is the ball’s height, measured in feet. When will the ball hit the ground? Round your answer to the nearest tenth.

Answer 1

The ball will hit the ground when h(t)=0 so:

-16t^2+10t+100=0

Using the quadratic equation:

t=(-10±√6500)/-32, and since t>0

t≈2.8 seconds (to nearest tenth of a second)

Answer 2

The ball will hit at 2.83 seconds.

Explanation:

It is given that,

Velocity with which a ball is thrown up is, v = 10 ft/s

Initial height, h = 100 ft

The equation for the height of an object as a function of t is given by :

`h(t)=-16t^2+10t+100`...............(1)

Where

t is the time in seconds

h is the ball's height measured in feet

We have to find the time when the ball hits the ground. It can be calculated by solving the quadratic equation of equation (1). On solving we get the value of t as :

t = 2.832 seconds

So, at 2.83 seconds the ball will hit the ground. Hence, this is the required solution.