Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

A ball is thrown straight up in the air at a velocity
of 100 feet per second. The height of the ball at t seconds
is given by the formula: h = 100t – 5t^2 Find the
maximum height of the ball and when it will land. Maximum Height ____ ft. Land after: ___ seconds


Sagot :

tochinwachukwu33

Answer:

Maximum height = 500 feet

Land after = 20 seconds

Step-by-step explanation:

Let us follow the movement of the ball as it is thrown upwards.  We notice that we get to a point, where the change in the height of the ball with respect to time becomes = 0. This means that the ball is at its maximum position.

Hence, mathematically, at maximum height, [tex]dh/dt = 0[/tex]

Recall,

[tex]h = 100t - 5t^2[/tex]

[tex]dh/dt=100-10t[/tex]

From our problem, we can also deduce that at the maximum height, the velocity of the ball is = 0. i.e [tex]dh/dt = 0[/tex]

hence,

[tex]0 = 100-10t\\t =10 seconds[/tex]

The time the ball takes to reach a maximum height is 10 seconds.

The next step will be to substitute this time value into the equation for the height.

[tex]h = 100(10) -5(10)^2=500 feet[/tex]

The maximum height is 500 feet

The object will land after t X 2 seconds = 10 X 2 = 20 seconds. This is because it will have to go up and come back down. This should take twice the time.

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.