A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor.(adsbygoogle = window.adsbygoogle || []).push({});

What horizontal distance does the ball cover before bouncing?

I keep getting 2.4m as my answer, but it's wrong!

Here's what I did (Please correct any mistakes):

4.6sin(15) = initial vertical velocity = 1.19m/s

1.19^2 = 2gh

1.42 = 2gh

1.42/(2*9.81)=h= .072m

upward time = 1.19 m/s - gt = 0

1.19/9.81 = t = .121sec

total height = .072 + 0.8 = .872m

0.872 = 1/2gt^2

0.872/4.9 = t^2 = 0.178, sqrt = t = 0.422sec

total time = 0.121 + 0.422= 0.543sec

horizontal velocity = 4.6cos(15) = 4.44m/s

4.44x 0.543 = 2.41m (answer)

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# Homework Help: Please Help Me Find Horizontal Distance?

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