Physics question about a ball thrown upward?

Question

An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

Can someone please explain HOW to do this? I don't just want the answer. Is there an equation I need to use orr .. ? I've been trying to figure this out and can't seem to get a correct answer :(

Any help is greatly appreciated!

Answer 1

This is essentially the same as the case where the object is launced from the ground with an initial velocity of 25 m/s. Calculate the hight for that problem and then multiply by 3.

So, v = v0 - g*t. Solve for t.

x = x0 + v0*t - (g*t^2)/2

In the modified problem x0 = 0 m and v0 = 25 m/s.

Solving for x gives the last 1/3 of the height for the original problem.

Answer 2

If the object reach 25 m/s after a time "t" then:

s = v0t - (1/2)gt^2 = (2/3)H

Where s = altitude at time "t" and H = the max altitude

If v0 is the initial velocity then the velocity v after some time t is:
v = v0 - gt = 25 and v0 = 25 + gt

So the altitude equation becomes:
s = 25t + (1/2)gt^2 = (2/3)H

Above 25m:
The velocity starts at 25 m/s and then drops to 0 at H. So if T is the time it takes to reach H from altitude s then:
v = v0 - gT = 25 - gT = 0 and T = 25/g

The total time to reach H is then T + t and the maximum equation for the maximum altitude can be written as:
H = v0(T + t) - (1/2)g(T + t)^2

H = (25 + gt)(25/g + t) - (1/2)g(25/g + t)^2
H = 25t + 625/g + 25t + gt^2 -(1/2)g(625/g^2 + 50t/g + t^2)
H = 25t + (1/2)gt^2 +[625/g -(625/2)/g]
H = 25t + (1/2)gt^2 + [(625/2)/g]

Using the equation for the initial altitude derived above:
25t + (1/2)gt^2 = (2/3)H

Ans substitute it into the equation for H:
H = (2/3)H + [(625/2)/g]
(H/3) = 95.66 meters

Maximum altitude is 95.66 meters.

Check:
(2/3)H = (2/3)95.66 = 63.77
s = 25t + (1/2)gt^2 = 63.77
4.9t^2 + 25t - 63.77 = 0
t = (-25 +/- SQRT(625 + 4*4.9*63.77))/9.8
t = 1.867 seconds

v0 = 25 + gt
v0 = 43.3 m/s

T = 25/g = 2.55 seconds
Total time to maximum = 2.55 + 1.867 = 4.418 seconds
H = (1/2)g(4.418)^2 = 95.66 meters
v0 = g(4.418) = 43.4 m/s

Answer 3

Physics? ball thrown upward vertically, optimal peak? a ball is thrown upward with an preliminary speed of 29.4 m/s. What optimal peak does the ball attain? whilst an merchandise is thrown upward, it decelerates 9.8 m/s each and each 2d, till its speed = 0 m/s. right now the object is at its optimal factor. Then it speeds up downward at 9.8 m/s each and each 2d. If the commencing up factor and end factor are the comparable, the path is symmetrical. the object strikes down the comparable distance it moved up. the desirable speed on the top of the downward trip equals the preliminary speed of the upward trip. The time to be triumphant in the pass as much as the optimal factor is the comparable using fact the time to fall backpedal to the commencing up factor. the fee of a ball with preliminary vertical speed of 29.4 m/s will cut back 9.8 m/s each and each 2d till the vertical speed = 0 m/s vf = vi + a*t vf = 0 vi = 29.4 a = -9.8 0 = 29.4 +(-9.8 * t) t = 3 seconds using fact the fee is reducing at a persevering with fee, the gap the ball strikes equals its effortless speed circumstances the time. effortless speed = (preliminary speed + very final speed) ÷ 2 effortless speed = (+29.4 + 0) ÷ 2 = 14.7 m/s Time = 3 seconds Distance = 14.7 m/s * 3 s = 40 4.a million m is it 44m or 88 m (40 4) are you able to apply d=vt NO, using fact the fee became no longer 29.4 m/s each and all the time!! v oftentimes means consistent speed. whilst an merchandise is shifting upward, its speed is reducing using gorgeous rigidity between the object and the earth. are you able to locate t? confident, by way of dividing the preliminary speed by way of the deceleration

Answer 4

Divide the maximum height by two thirds. Divide this by 4.9 and find the square root of this number, this will give the time in seconds that the ball is in motion, multiply this by 9.8.