Asked by Jay
A ball is thrown from a cliff upward with a speed of 27 m/s and at an angle of 71 degrees to the horizontal. How high does it go above the cliff to the nearest tenth of a meter?
answer is 33.3.
I am using 27sin71.
answer is 33.3.
I am using 27sin71.
Answers
Answered by
Devron
That is the initial velocity in the y-direction.
Use the following formula:
Vf^2=Vi^2+2gd
Where
d=?
Vi=(27m/s*Sin71º)+2(-9.8m/s^2)d
Vf=0
g=-9.8m/s^2
Solve for d,
0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d
0=651.7m/s-19.6m/s^2*d
(-651.7m/s/-19.6m/s^2)=d
d=33.25m=33.3m
Use the following formula:
Vf^2=Vi^2+2gd
Where
d=?
Vi=(27m/s*Sin71º)+2(-9.8m/s^2)d
Vf=0
g=-9.8m/s^2
Solve for d,
0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d
0=651.7m/s-19.6m/s^2*d
(-651.7m/s/-19.6m/s^2)=d
d=33.25m=33.3m
Answered by
Devron
That is the initial velocity in the y-direction.
Use the following formula:
Vf^2=Vi^2+2gd
Where
d=?
Vi=(27m/s*Sin71º)
Vf=0
g=-9.8m/s^2
Solve for d,
0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d
0=651.7m/s-19.6m/s^2*d
(-651.7m/s/-19.6m/s^2)=d
d=33.25m=33.3m
Use the following formula:
Vf^2=Vi^2+2gd
Where
d=?
Vi=(27m/s*Sin71º)
Vf=0
g=-9.8m/s^2
Solve for d,
0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d
0=651.7m/s-19.6m/s^2*d
(-651.7m/s/-19.6m/s^2)=d
d=33.25m=33.3m
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.