Vertical Motion under Gravity Q)

[Lavace]

Homework Statement

Apologies for so many posts, however this is the last question.

A stone his thrown vertically upwards with speed 16m/s from point h meters above the ground. The stone hits the ground 4s later.
Find the value of h.

The Attempt at a Solution

I really haven't a clue on how to start this question, I haven't really encoutered this before, and would end up doing my own sledgehammer way.
But otherwise, can someone people give hints on how to solve this, I know all the kinematic equations, its just this situation confuses me cause of h.

Thanks in advance.
Lav.

 

[skeeter]

delta y = (voy)t - 4.9t^2

when the stone hits the ground, its vertical displacement will be the opposite of its initial height ...

delta y = -h

-h = 16t - 4.9t^2

sub in t = 4 ... h = ?

 

[m2003]

I would first start by defining the problem and identifying the known and unknown variables. In this case, we know the initial velocity (16m/s), the time (4s), and the acceleration due to gravity (9.8m/s^2). The unknown variable is the initial height (h) above the ground.

Next, I would use the kinematic equations to solve for h. The equation we need is h = h0 + v0t + 1/2at^2, where h0 is the initial height, v0 is the initial velocity, t is the time, and a is the acceleration. We can rearrange this equation to solve for h: h = h0 + v0t - 1/2at^2.

Plugging in the known values, we get h = h0 + (16m/s)(4s) - 1/2(9.8m/s^2)(4s)^2. Simplifying, we get h = h0 + 64m - 78.4m. Since we know the stone hits the ground (h = 0) after 4 seconds, we can set the equation equal to 0 and solve for h0. 0 = h0 + 64m - 78.4m. Therefore, h0 = 14.4m.

Therefore, the initial height of the stone above the ground was 14.4 meters.