a stone thrown from the top of a building is given an initial velocity of 20.0 m

Question

a stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. the stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down

a) using t=0 s as the time the stone leaves the thrower's hand at position, determine the time at which the stone reaches its maximum height
b) find the maximum height of the stone
c) determine the velocity of the stone when it returns to the height from which it was thrown
d) find the position of the stone at t= 5.00 s

Explanation / Answer

to find the time the stone is at its max hieght, we will find the time it takes to go up and then come back down to the hieght its thrown from (or, when the change in x = 0), so we get the equation...

0=vo*t+.5*a*t^2

0=20t-.49t^2

0=t(20-.49t)

t = 0, 4.08 s <--- we're only interested in the non-zero answer

now what we'll do is divide that by 2, since that will get us half of the parabola, or the vertex

4.08/2 = 2.04 s

now, the max hieght of the stone we'll find by realizing the v=0 at the peak.

v^2 = vo^2+2ax

0 = 4-19.6x

400 = 19.6x

400/19.6 = x = 20.41 m

thats just the distance it goes from where it is thrown though so we'll add the original hieght to this...

20.41+50 = 70.41 m

intuitively we can guess that the velocity will be the same when it gets back to the same hieght it was thrown at, just in the opposite direction, and the math supports this...

v^2 = vo^2+2ax

v^2 = 20^2+0

v=20 m/s

and finally the position of the rock after 5 s, we just plug in the numbers...

x = vo*t+.5*a*t^2

x = 20*5-4.9*25

x = -22.5 m

and add this to the original 50 m to get...

50-22.5 = 27.5 m

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