A rock is dropped from a treetop 19.6 m high, and then,
1.00 s later, a second rock is thrown down. With what
initial velocity must the second rock be thrown if it is to
reach the ground at the same time as the first?
I'm not sure how to find the initial velocity of the thrown rock, since it's not 0
What I have so far:
1st rock:
Yo=o
Yf=19.6 m
Vo=o
Vf=
T=
A= -9.80 m/s^2
I think I've found the final velocity and time of the 1st rock
(Vf^2=Vo^2-2g(Y-Yo)
or final velocity squared = initial velocity(0 m/s)^2 -2(9.80)(-19.6-0)
which is = to -19.6 m/s
Time = 2 seconds
(Vf=Vo-gt, Vf/Vo-g=T, -19.6 m/s /-9.80 m/s^2, 2 seconds
How do I find the initial velocity of a throw object when I know:
Yo=o
Yf=19.6 m
T= 2-1 = 1 second
A- /9.80 m/s^2
Thanks!
Wess, there are 2 cases in this prob:
Case 1
U nid to find how long the first rock took to hit the ground: use the formular- s=ut+0.5gt^2 where s=19.6m, u=0, t=?
I gor t=1.97s
Case 2
The second rock took 1.97s-1s to hit the grnd, use same formula & find u
Thank you very much!
To find the initial velocity of the second rock, you can use the equation of motion for a vertically thrown object. The equation is:
Yf = Yo + Vo*t + (1/2)*a*t^2
Where:
- Yf is the final position (19.6 m in this case),
- Yo is the initial position (0 m),
- Vo is the initial velocity (what we want to find),
- t is the time (1 second in this case),
- a is the acceleration due to gravity (-9.8 m/s^2).
You can rearrange the equation for Vo:
Vo = (Yf - Yo - (1/2)*a*t^2) / t
Substituting the given values:
Vo = (19.6 m - 0 m - (1/2)*(-9.8 m/s^2)*(1 s)^2) / (1 s)
Simplifying the equation:
Vo = (19.6 m + 4.9 m) / 1 s
Vo = 24.5 m/s
Therefore, the second rock must be thrown with an initial velocity of 24.5 m/s in order to reach the ground at the same time as the first rock.
To find the initial velocity of the second rock, you can use the equation of motion for the vertical displacement:
Y = Yo + Vot + (1/2)at^2
where:
- Y is the final position (in this case, 19.6 m)
- Yo is the initial position (in this case, 0 m)
- Vo is the initial velocity (unknown)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken (in this case, 1 second)
Plugging in the values, we get:
19.6 = 0 + Vo(1) + (1/2)(-9.8)(1^2)
Simplifying the equation:
19.6 = Vo - 4.9
Vo = 19.6 + 4.9
Vo = 24.5 m/s
Therefore, the initial velocity of the second rock must be 24.5 m/s to reach the ground at the same time as the first rock.