Asked by Jay

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.

Answers

Answered by Steve
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
Answered by Jay
the answer to the question is 58 why is it not matching up?
Answered by Steve
Ahh. I used the horizontal component of v, not the vertical. Chane cos to sin and things should work out.

Shoulda caught that, guy...
Answered by Jay
Thank you
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions