Question
Use the formula for the height h of an object that is traveling vertically (subject only to gravity) at time $t$ : $$h=-16 t^{2}+v_{0} t+h_{0}$$where $h_{0}$ is the initial height and $v_{0}$ is the initial velocity; $t$ is measured in seconds and h in feet.A ball is thrown upward from the top of a 96 -foot-high tower with an initial velocity of 80 feet per second. When does the ball reach its maximum height and how high is it at that time?
Step 1
We know that the initial height $h_{0}$ is 96 feet, the initial velocity $v_{0}$ is 80 feet per second, and the acceleration due to gravity is -16 feet per second squared. So, we can write the equation as: \[h=-16t^{2}+80t+96\] Show more…
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Use the formula for the height h of an object that is traveling vertically (subject only to gravity) at time $t$ : $$h=-16 t^{2}+v_{0} t+h_{0}$$ where $h_{0}$ is the initial height and $v_{0}$ is the initial velocity; $t$ is measured in seconds and h in feet. A ball is thrown upward from the top of a 96 -foot-high tower with an initial velocity of 80 feet per second. When does the ball reach its maximum height and how high is it at that time?
You will need the formula for the height $h$ of an object above the ground at time $t$ seconds: $$h=-16 t^{2}+v_{0} t+h_{0}$$ this formula was explained on page 249 A ball is thrown straight up from a 40 -foot-high tower with an initial velocity of 56 feet per second. (a) During what time interval is the ball at least 8 feet above the ground? (b) During what time interval is the ball between 53 feet and 80 feet above the ground?
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