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
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...
Shoulda caught that, guy...
Answered by
Jay
Thank you