Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.