A ball is launched at an angle of 59.5 degrees up from the horizontal, with a mu

Question

A ball is launched at an angle of 59.5 degrees up from the horizontal, with a muzzle velocity of 6.5 meters per second, from a launch point which is 1.2 meters above the floor. How far horizontally (in meters) from the launcher will it land on the floor? Use 9.82 meters per second for "g".  

If you were asked to find the horizontal distance from the launcher when it is some distance ABOVE the floor (instead of when it HITS the floor), what about your previous equations/calculation would you have to change?

Explanation / Answer

H ( horizobtal distance) =( 6.5 cos 59.5/ 9.8 ) ( 6.5 sin 59.5 + sqroot [ 6.5 ^2 sin ^2 59.5 + 2 x 9.8 x 1.2] )

= 0.3366 ( 5.6 + 7.408 )

= 4.378 m apprx

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