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?

Question

answer is 33.3.

I am using 27sin71.

Answer 1

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

Answer 2

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