The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y = 8t – 5t2 and x = 6t. If x and y are measured in metres, the velocity of projection is

x=6t⇒x=aty=8t−5t2⇒y=bt−ct2u=a2+b2=62+82=10ms−1

The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y = 8t – 5t2 and x = 6t. If x and y are measured in metres, the velocity of projection is

x=6t⇒x=aty=8t−5t2⇒y=bt−ct2u=a2+b2=62+82=10ms−1

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The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y = 8t – 5t² and x = 6t. If x and y are measured in metres, the velocity of projection is

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8 ms^-1

6 ms^-1

14 ms^-1

10 ms^-1

answer is D.

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Detailed Solution

$x = 6 t \Rightarrow x = at$
$y = 8 t - 5 t^{2} \Rightarrow y = bt - {ct}^{2}$
$u = \sqrt{a^{2} + b^{2}} = \sqrt{6^{2} + 8^{2}} = 10 {ms}^{- 1}$

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The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y = 8t – 5t2 and x = 6t. If x and y are measured in metres, the velocity of projection is

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The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y = 8t – 5t² and x = 6t. If x and y are measured in metres, the velocity of projection is