SOLUTION: *CAN YOU PLEASE EXPLAIN HOW YOU GOT THIS ANSWER FOR STUDYING PURPOSES*
Problem: A ball is thrown vertically upward into the air. If the ball started from a height of 101 feet of
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Question 1158668: *CAN YOU PLEASE EXPLAIN HOW YOU GOT THIS ANSWER FOR STUDYING PURPOSES*
Problem: A ball is thrown vertically upward into the air. If the ball started from a height of 101 feet off the ground, use the following formula to represent the projection of the ball
H=-16t^2+38t+101
where H is the height of the ball after t seconds have passed. Based on this information, answer the following questions.
1)When does the ball reach its maximum height?
Your answer must be expressed as a decimal rounded to 2 decimal places with correct units. Hint: The maximum occurs at the vertex of the parabola.
2)What is the maximum height of the ball?
Your answer must be expressed as a decimal rounded to 2 decimal places with correct units.
3)When does the ball return to the ground?
You must use the quadratic formula to solve this problem. Your answer must be expressed as a decimal rounded to 2 decimal places with correct units.
Hint: The ball reaches the ground when the height equals zero.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
H=-16t^2+38t+101
It helps to draw this.
The value of t at the highest point is where t=-b/2a, and b is 38 and a=-16
That will occur when t=-38/-32 seconds or 1.1875 or 1.19 seconds
Put that value of t bad into the equation
-16(1.1875^2)+38(1.1875)+101=-22.5625+45.125+101=123.5625 or 123.56 feet
set the equation = to 0 and solve using the quadratic formula. I will rewrite with a positive coefficient of the first term: 16t^2-38t-101=0
t=(1/32)(38+/- sqrt 1444+6464; sqrt term =88.93
the positive root is t=(126.93/32)=3.97 seconds
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