soccerball3211
Junior Member
- Joined
- May 16, 2005
- Messages
- 84
A stone is projected at a cliff of height h with an initial speed of 49.0 m/s directed at an angle 0 = 58.0° above the horizontal. The stone strikes at A, 5.48 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.
First I found the components of the projectile. I used the equations V sub x = V sub i cos theta and V sub y = V sub i sin theta.
Using these formulas I found the x component to be 25.966m/s and the y component to be 41.554 m/s.
A.) I then used the eqation y = (V sub y)t - (1/2)gt^2
I found the height of the cliff to be 80.6 m
C.) I then figured out part c by using the eqation t = (V sub i sin theta)/g
This gave me 4.24s.
To find max height I used d sub y = v sub y (t)
This gave me 176.2 m for the max height
B.) For part B i used the equation V sub y = g(t final-t max)
=-9.8m/s^2(5.48-4.24)
This gave me -12.12 m/s
Can someone check my work?
First I found the components of the projectile. I used the equations V sub x = V sub i cos theta and V sub y = V sub i sin theta.
Using these formulas I found the x component to be 25.966m/s and the y component to be 41.554 m/s.
A.) I then used the eqation y = (V sub y)t - (1/2)gt^2
I found the height of the cliff to be 80.6 m
C.) I then figured out part c by using the eqation t = (V sub i sin theta)/g
This gave me 4.24s.
To find max height I used d sub y = v sub y (t)
This gave me 176.2 m for the max height
B.) For part B i used the equation V sub y = g(t final-t max)
=-9.8m/s^2(5.48-4.24)
This gave me -12.12 m/s
Can someone check my work?