Question

Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at speed v and it spends time t in the air before smashing into the ground. Not to be outdone, Guy then throws his watermelon at speed 2v. How long is Guy’s watermelon in the air before it smashes back into the ground?

Answer 1

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

$v_f=u_i+at'$

0=v-gt'

$t'=\frac{v}{g}$

and time to reach bottom is also t'

t=2t'

$t=2\frac{v}{g}$

For guy throws his watermelon at speed 2v

So total time in air will be

$t''=2\frac{2v}{g}$

$t''=4\frac{v}{g}$

t''=2t