If a ball is shot from the ground straight up into the air with an initial veloc
ID: 3135959 • Letter: I
Question
If a ball is shot from the ground straight up into the air with an initial velocity of 64 ft/sec its height in feet in feet after t seconds is given by h(t)=64t-16t^2 When will the ball hit the ground, and what is the maximum height the ball will reach? If a ball is shot from the ground straight up into the air with an initial velocity of 64 ft/sec its height in feet in feet after t seconds is given by h(t)=64t-16t^2 When will the ball hit the ground, and what is the maximum height the ball will reach? When will the ball hit the ground, and what is the maximum height the ball will reach?Explanation / Answer
Given h(t)=64t-16t^2
derive it:
h`(t)=6432t the values for t represent when the height of the ball is 0.
set derivative equals to 0
t=64/32
t=2 sec
after 2sec the ball will return the ground.
maximum height:
use this value of t into your trajectory:
h(2)=hmax642164=64 feet