Question 413427: An object is thrown upward from a height of 50 ft with an initial velocity of 48 ft per sec. The height in feet after t seconds is given by h=-16t^2+48t+50
a. Find the height of the object 3 sec after it is thrown up
b. Find the maximum height of the object and the time it takes the object to reach this height
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! An object is thrown upward from a height of 50 ft with an initial velocity of 48 ft per sec. The height in feet after t seconds is given by h=-16t^2+48t+50
a. Find the height of the object 3 sec after it is thrown up
b. Find the maximum height of the object and the time it takes the object to reach this height
..
h=-16t^2+48t+50
after 3 seconds,
a. h(3)=-16*3^2+48*3+50
=-144+144+50=50
What happened is that the object reached its peak then returned to its original height after 3 seconds.
b.You can use the formula, -b/2a to find the time object reaches its maximum height. a=-16, b=48,
Time to reach maximum height =-b/2a=-48/2*-16=-48/-32=3/2 seconds
Maximum height = -16(3/2)^2+(3/2)*48+50=-36+72+50=86 feet
ans:
3 seconds after object is thrown upward, it returns to a height of 50 feet going down.
It takes 1.5 seconds for object to reach its peak after thrown upward.
Maximum height object will reach=86 feet
|
|
|
