Posted by **asem** on Friday, March 16, 2012 at 7:25am.

baseball is hit with a speed of 30.0 m/s at an angle of 50 It lands on the flat roof of a 10m tall nearby building, If the ball was hit when it was 1.5 m above the ground, what horizontal distance does it travel before it lands on the building?

- physical -
**asem**, Friday, March 16, 2012 at 7:49am
V 0x = 30*cos 50 = 19.28 m/s

V 0y = 30*sine 50 = 22.98

Y = 14-1.5 =12.5

Y = V 0y*t -0.5*g*t^2

12.5 = 22.98*t -4.9*t^2

Solve the quadratic equation for t. t = 4.06 seconds

X = v 0x*t

X = 19.28*4.06 = 78.27 meters

- physical -
**asem**, Friday, March 16, 2012 at 7:50am
The ball travels in the vertical direction with initial vertical component of 30 sin 50 [22.98 m/s^2] upward till its velocity component in vertical direction becomes let us say V. The final vertical component after travelling [14 - 1.5] = 12.5 m is given by

[30 sin 50]^2 - V^2 = 2x9.8x12.5 or

V^2 = 528.14 - 245 = 283.14 or V = 16.826 m/s

Time required to reduce the vertical component from 22.98 to 16.82 would be 6.16/9.8 s = 0.628 s

The horizontal distance travelled by the ball in this time is

30 cos 50 x 0.628 = 12.11 m

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