Distance and time

[12345678]

A particle is projected vertically upwards from ground levelwith a velocity of 50ms^-1. For how long will it be more than 70m above ground?
So my method of thinking was to find the particles peak height.
We know U (initial velocity) = 50ms
Acceleration due to gravity is 9.81N/Kg
At the peak height, final velocity V = 0
Using the equation V² = U² + 2as
S (distance) = V² - U² / 2a
= 0 - 50² / 2(-9.81) = 127.41m
I then found the timeit took to reach this
S = (U+V)/2 X t
S = 50 + 0 / 2 * t
127.41m = 25t
T = 5.1 seconds
I then did time taken to reach 70m…
70 = 25t
T = 2.8 seconds
Thus the time taken from 70m to peak height = 5.1 – 2.8 =2.3 seconds
As the ball falls in a symmetrical pattern, the time above70m = 2.3 x 2 = 4.6 seconds.
However, the answer is 6.82 seconds- anybody know why?

 

[Quaid]

As the exercise asks about height, I think the formula for height may be more direct.

h = h_o + v_o t - 4.9t^2

where h_o is the initial height and v_o is the initial velocity (and -4.9 comes from 1/2*g)

Set h equal to 70, and solve for the two times. Their difference is the answer.

Cheers :cool:

PS: The answer you posted (6.82 seconds) seems improperly rounded

 

[12345678]

As the exercise asks about height, I think the formula for height may be more direct.

h = h_o + v_o t - 4.9t^2

where h_o is the initial height and v_o is the initial velocity (and -4.9 comes from 1/2*g)

Set h equal to 70, and solve for the two times. Their difference is the answer.

Cheers :cool:

PS: The answer you posted (6.82 seconds) seems improperly rounded

Cheers for the reply. You wouldn't know how to get this answer or formula using suvat equations would you? (Because that equation isn't on my AS physics equation sheet)

 

[Quaid]

how to get this answer or formula using suvat equations

The formula for height (also known as position) is a suvat equation.

See the reference page HERE.

Instead of using symbols h, v_o, and g, they use s, u, and a (respectively).