A ball is thrown straight upward and returns to the thrower\'s handafter 2.80 s

Question

A ball is thrown straight upward and returns to the thrower's handafter 2.80 s in the air. Asecond ball is thrown at an angle of 40.0° with thehorizontal. At what speed must the second ball be thrown so that itreaches the same height as the one thrown vertically?
I have triedseveral times to come up with the answer but it comes out wrongeach time. Any help would be very appreciated!
I have triedseveral times to come up with the answer but it comes out wrongeach time. Any help would be very appreciated!

Explanation / Answer

let the speed of the one throw be v1, total in air time t1 is: t1 = 2*v1/g = 2.80s =>v1 = 2.80s*9.8m/s2 / 2 = 13.72 m/s the max heigh H = v12 / 2g = (13.72m/s)2/ 2*9.8m/s2 = 9.60m let the speed of the second throw be v2, we have: v2y = v2*sin(40.0) the max heigh H = 9.60m = (v2y)2 / 2g =[v22 sin2(40.0)/2g] =>v2 = [2*9.8m/s2 *9.60m /sin2(40.0)] = 21.34 m/s second throw need to be have speed 21.34m/s to reach the sameheigh as first throw. .

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