Question

If the time of flight of a projectile is doubled what happens to the maximum height attained

Answer 1

As we know that the max. height is,
h=u^2sin^2theta/2g
We can also write it as
»h=(4u^2sin^2theta/8g^2 )*g
»h=T^2 *g/8 [T=2u sintheta/g]

As we can see from the above Eq. that the height(max) is directly proportional to the sqare of time of flight (T) and when we double the time of flight the max. height becomes four times the time of flight.

Hence the ans is four times the time of flight.

Answer 2

Answer:

height becomes 4 times

Explanation:

H = $\frac{u^{2}sin^{2} }{2g}$

multiplying 4*g to both numerator and denominator we get,

$H =[ \frac{4u^{2}sin^{2} }{8g^{2} } ]*g$

we know that,

$T = \frac{2u*sin}{g}$

substituting T in H we get,

$H = \frac {T^{2} }{8} *g$