Question

A stone is thrown vertically upwards with a speed of 28 m/s. How fast is it moving when it reaches a height of 28.0 m? How much time is required to reach this height?

Answer 1

This is what I have so far: d = 28.0m, a = -9.8m/s(since its thrown upwards), initial velocity is 28.m/s. The unknown is the final velocity.For my equation, I figured i would use this one: vf^2 = vi^2+2g*d, where vf is final velocity, vi is initial velocity, g is gravity or acceleration and d is the distance
My final answer for the first part was: vf^2=(28.0m/s)^2 + 2(-9.8m/s)(28m) -784m/s^2/-548.9 = 1.43

Answer 2

how about
v= 28 - g*t
and distance d= 28t- (1/2) g *t^2

Answer 3

so its d=(28m/s)t+1/2(-9.8m/s)t^2?

Answer 4

with d=28 m
solve for t to get the time. you get 2 answers (the first when the stone is going up, the second when the stone is falling)

Answer 5

okay so: 20m=28m/st+1/2(-9.8m's^2)t^2

28m=-23.1
=1.21

Answer 6

where did you get 20m? isn't it 28?
you have to solve

4.9 t^2 -28t +28 = 0

Answer 7

ops sorry, I meant 28.

Answer 8

4.9 *1.21*1.21 -28*1.21+28= 1.29 which is not zero

you must use t= (-b ± sqrt(b*b-4*a*c)/(2a)
where a= 4.9, b= -28, c=28

Answer 9

ah I see quadratic formula. Thanks.

Answer 10

I got 4.42 and 1.29 for time, but the answers weren't right. I've plugged in all the values for the formula.

Answer 11

I get t= 1.2922 seconds for when the stone first reaches a height of 28 m
the velocity at this time is
v= 28 - 9.8t= 15.336 m/s