1.

The height y and the distance x along the horizontal plane of a projectile 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 :

Answer»

`6ms^(-1)`
`8ms^(-1)`
`10ms^(-1)`
`14ms^(-1)`

SOLUTION :`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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