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
    
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