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The height y and the distance x along th...

The height y and the distance x along the horizontal plane of a prohectile on a certain planet (with no surrounding atmosphere) are given by y=8t-`5t^(2)` meter and x=6t meter,where t is in seconds.The velocity with which the projectile is projected is
(Acceleration due to gravity =9.8 m `s^(-2)`)

A

`6ms^(-1)`

B

`8ms^(-1)`

C

`10ms^(-1)`

D

`14ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`y=8t-5t^(2):.v_(y)(dy)/(dt)=8-10t`
At t=0, `v_(y)=8m//s`
Similarly `x=6t:v_(x)=(dx)/(dt)=6m//s`
Now `v=sqrt(8^(2)+6^(2))=10m//s`
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