A baseball is thrown off a cliff at an angle of 57 degrees with a speed of 25 m/s. How high is the cliff to the nearest tenth of a meter, if the ball hits the ground 6.2 sec later.

Answer 1

y = h + 25cos57 t - 4.9t^2
So, if takes 6.2 sec to hit at y=0, then

h + 13.6*6.2 - 4.9*6.2^2 = 0
h = 104.0m

Answer 2

the answer to the question is 58 why is it not matching up?

Answer 3

Ahh. I used the horizontal component of v, not the vertical. Chane cos to sin and things should work out.

Shoulda caught that, guy...

Answer 4

Thank you