Can u help me with this physics problem?

Question

You have to use this formula

h(t) = -4.9^2 + v subzero t+ h sub zero

v subzero t = v sub zero times T

vsub zero = innitial upward velocityy
h sub zero = meters above the ground
t = seconds
h(t) = meters aboce the ground

A stone is thrown with an upward velocity of 14m/s from a cliff 30 m high.

When i plug in all teh stuff it comes out to.....

h(t) = -4.9t^2 + 14t + 30

How can i tell when the stone reaches its highest elevation?

Answer 1

okay here's how it goes,

You know there are 2 instances when the stone would be at the height of 30 meters, the one time when it is starting, t=0 seconds and the second time when it falls from the highest point and pass this same mark on its way down the cliff.

so essentially you are looking at solving the equation
-4.9t^2 + 14t = 0 (note this is an abstraction of the equation you have above) - which means you are trying to determine the values of t whereby h(t) = 30 meters.

from this equation you get
(t) x (-4.9t +14) = 0
which yields 2 possibilities of t,
first t = 0 we already know, the second t = 14/4.9 = approximately 2.857 seconds.

now this is not the final answer since this is the time the stone passes this point, you'd need to divide it by half to determine the time it is at its peak (since time taken to go up should be equal to time it takes to fall down to the same height).

hope it helps....

Answer 2

h(t) = -4.9t^2 + 14t + 30
V(t) = 9.8t + 14

You've got the initial equation right.

To find the highest point, you have to find out when the upward velocity is 0, because V=0 when the acceleration due to gravity has cancelled the initial upward velocity (14t). After that point, the stone begins to fall as gravity has overtaken your initial throw of the stone.

To find v(t), you take the derivative of h(t). To do the derivative, take each part of the equation, multiply the coefficient by t's exponent, and decrease the exponent by 1:
The derivative of 30 is 0
The derivative of 14t is 14 (14 * 1(t^0)
The derivative of -4.9t^2 is 9.8t (-4.9 * 2 (t^(2-1))

Therefore V(t) = -9.8t + 14

Since we're looking for v(t) = 0…
0 = -9.8t + 14
14 = 9.8t
T = 1.43 sec

Check:
h(1.4 sec) = -9.604 + 19.6 + 30 = 39.996
h(1.43 sec) = -10.02 + 20.02 + 30 = 40m (the highest point).
h(1.5 sec) = -11.025 + 21 + 30 = 39.975