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)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Please Help Me Find Horizontal Distance?

**Physics Forums | Science Articles, Homework Help, Discussion**