If object is thrown upward with an intitial velocity of 15 ft per second, then its height is given by h=-15t^2 + 60t. What is the maximum height?
Two ways to find it.
1) AT the top, max height, will be one half of the time it would come back down and hit the earth at h=0
0=-15t^2+60t solve for t.
Then divide that by two.
2) At max height, the change in height/per time is zero.
dh/dt= -30t+60
set that to zero, and solve for t.
If you don't know derivatives yet, forget this method.
If object is thrown upward with an intitial velocity of 15 ft per second, then its height is given by h=-15t^2 + 60t. What is the maximum height?
The time to reach maximum height is given by Vf = Vo - gt where Vf = the final velocity, Vf = 0, Vo = the initial velocity, g = acceleration due to gravity, ~32), and t = the time.
Therefore, 0 = 15 - 32t making t = .46876 sec. sec.
The height reached is given by h = Vot - 16t^2. You give an equation of h = 60t - 15t^2 implying that Vo = 60, not 15, and g = 30, not 32.
Using your equation, h = 60(.46875) - 15(.46875)^2 = 24.829 ft. ft.
Using the initial velocity stated, h = 15(.46875) - 15(.46875)^2 = 3.735 feet.
If the real initial velocity is 15 fps, then 60 does not belong in the equation for height, rather 15 instead..
I believe this result derives from using the wrong value for g which should be 16.
Using g == 16 and the correct V0 = 15 in the equation for height, we get
0 = 15 - 16t making t = ..9375 sec.
Then, h = 15(.9375) - 16(.9375)^2 = 14.0625 - 14.0625 = 0.
This leads to h = 15(1) - 15(1) = 0
Therefore, with Vo = =15
To find the maximum height using the first method:
1) We want to find the time it takes for the object to reach maximum height, which is also the time it takes to come back down and hit the ground at h=0.
Setting h=0 in the equation h = -15t^2 + 60t:
0 = -15t^2 + 60t
Rearranging the equation, we get:
15t^2 = 60t
Dividing both sides by 15 gives:
t^2 = 4t
Subtracting 4t from both sides gives:
t^2 - 4t = 0
Factoring out t from both terms gives:
t(t - 4) = 0
So either t = 0 or (t - 4) = 0.
Therefore, t = 0 or t = 4.
2) Since we are looking for the time it takes to reach maximum height, we divide t by 2:
t/2 = 4/2
t/2 = 2
So the time it takes for the object to reach maximum height is 2 seconds.
To find the maximum height using the second method:
1) Find the derivative of the height equation with respect to time:
h' = -30t + 60
2) Set the derivative equal to zero and solve for t:
-30t + 60 = 0
-30t = -60
t = 2
So the time it takes for the object to reach maximum height is again 2 seconds.
Using either method, we find that the maximum height is reached after 2 seconds.
To find the maximum height, we can use either of the two methods mentioned:
Method 1:
1) Find the time it takes for the object to reach its maximum height and come back down to the ground. To do this, we set the height equation h = -15t^2 + 60t equal to 0 since the object is at the ground level. So, -15t^2 + 60t = 0. Solve this equation for t.
2) Divide the time obtained in step 1 by 2 to get the time it takes for the object to reach its maximum height.
3) Plug the value of t/2 back into the height equation h = -15t^2 + 60t to find the maximum height.
Method 2:
1) Take the derivative of the height equation with respect to time, which gives us the rate of change of height with respect to time (dh/dt).
2) Set dh/dt equal to 0 because at the maximum height, the change in height per unit time is 0.
3) Solve the equation dh/dt = -30t + 60 for t to find the time it takes for the object to reach its maximum height.
4) Plug the value of t back into the height equation h = -15t^2 + 60t to find the maximum height.
Both methods will give you the same result for the maximum height.