Question

7. A projectile shot into air at some angle with the
horizontal has a range of 200m. If the time of
Night is 5s, then the horizontal component of
the velocity of the projectile at the highest point
oftrajectory is
1) 40ms!
2)Oms:
3) 9.8ms! 4) 20ms!​

Answer 1

Explanation:

Do u know the formula for range and time of flight... otherwise u can also derive that..

Answer 2

Answer:

$let \: the \: angle \: of \: projection \: be \: \alpha \\ and \: intial \: velocity \: u \: . \\ \\ given \\ \\ horizontal \: range \\ = \frac{2 {u}^{2} \sin( \alpha ) \cos( \alpha ) }{g} = 200 \\ \\ time \: of \: flight \\ = \frac{2u \sin( \alpha ) }{g} = 5$

We know that horizontal component of velocity remains unchanged .

$u \cos( \alpha ) = const.$

To get this just divide both equations

$\frac{2 {u}^{2} \sin( \alpha) \cos( \alpha ) }{2u \sin( \alpha ) } = \frac{200}{5} \\ \\ u \cos( \alpha ) = 40 \: \frac{m}{s}$