How fast is it moving when it reaches a height of 16.0m? How long is required to reach this height?

Question

A stone is thrown vertically upward with a speed of 20 m/s2.
Answer is + or - 9.30 m/s, 1.09 s, 2.99 s

How do u get this answer?
Please help
Thank You

Answer 1

Two equations to start with:

y = y0 + v0*t - (1/2)*9.8*t^2

and
y' = v = v0 - 9.8t

where t is time, y is position and v is velocity.

Find time first, using the 16.0 m..
Say y0 = 0 meters, in other words, the ground

16 = 0 + 20*t - (1/2)*9.8*t^2
0 = -16 + 20t - 4.9t^2

t = [ -20 +/- sqrt( (20^2) - (4)(-4.9)(-16) ) ] / (2)(-4.9)
t = 1.09 seconds or 2.99 seconds.

There are two times when the stone hits 16.0 meters, on the way up and on the way down. The question asks for the instance when the stone is on the way up, so use t = 1.09 seconds.

How fast is it moving?

Use the second equation for velocity:

y' = v = v0 - 9.8t
v = 20 - 9.8(1.09)
= +9.30 m/s, plus since it is in the direction opposite to the direction of the acceleration of gravity.

Answer 2

You said the initial speed was 20 m/s2. Wrong units for speed. I'm going to guess you meant Vo = 20 m/s.

The formula V^2 = Vo^2 + 2*a*y should work.
Use Vo = 20 m/s, a = -9.8 m/s^2, and y = 16 m. You should get V = 9.3 m/s.

How long to get up to 16 m? The formula y = (V + Vo) * t / 2 should give you 1.09 s.

Why does the answer have 4 numbers? Notice that the stone is still going up when it reaches 16 m. It will reach a peak height at 0 velocity and fall back down, passing 16 m on the way to the ground. The same a = -9.8 m/s^2 acts on it the whole way. The 2 halves of the path from 16 m -> peak -> 16 m will be mirror images of each other so the speed the second time it passes 16 m is -9.3 m/s.

Time between the 2 passes by 16 m? This time use the formula
V = Vo +a*t
but just consider the time from 16 m to the peak. V = 0 at the peak, Vo = 9.3 m/s, and a = -9.8 m/s^2.
You should get t = .95 s. Double that to get the time from 16 m -> peak -> 16 m and add to the time it took to get up to 16 m the first time (1.09 s). You should get 2.99 for that. Take a break.

Answer 3

the equation of the stone is :
h = -1/2 gt² + V0 t +h0
with g = 9.8 and v0 = 20
so h = - 4.9 t² + 20 t +h0
it reach 16m so h= 16
4.9 t² -20 t +16 = 0
so t1 = (20 - sqrt(400 - 4*16*4.9))/2/4.9 = 1.09s
t2 = (20 + sqrt(400 - 4*16*4.9))/2/4.9 = 2.99 s

velocity at this point is :
v = -gt +v0 = - 9.8 t +20
v1 = 9.30 m/s
v2 = -9.30 m/s

Answer 4

V^2= (Vo)^2 + 2gy
g= -9.8m/s
v^2= (20)^2 -(2)*(9.8)*(16)
V^2= 400 -313.6
v^2= 86.4
v=9.3m/s
V=Vo +at
9.3=20-9.8t
t= 1.19s