Question

the height y and the distance x along horizontal plane of a projectile on a certain planet with no surrounding atmosphere are given with y=8t -5t^2 and x =6t ,where t is in second .the velocity with which projectile is projected

Answer 1

HEY BUDDY

YOU GOT THIS ANSWER BY UDIT

velocity in x

dx/dt=6

velocity in y

dy/dt=8-10t

for initital velocity t=0


dy/dt=8

net velocity

$\sqrt{6{}^{2} +8 {}^{2} } = 10$

thanks

Answer 2

Refer The Attachment ⬆️

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