Question

if time to flight of a projectile is 10s and range 400m then maximum height attained will be

Answer 1

Given:

Time to flight of a projectile is 10s and range 400m.

To find:

Max height attained ?

Calculation:

Let initial velocity be "v" and the angle of projection be $\theta$:

$\therefore \: range = 400\:m$

$\implies \: \dfrac{ {v}^{2} \sin(2 \theta) }{g} = 400 \: \: \: ....(1)$

Now, time for Projectile:

$\therefore \: time = 10\:sec$

$\implies \: \dfrac{2v \sin( \theta) }{g} = 10$

$\implies \: \dfrac{v \sin( \theta) }{g} = 5$

$\implies \: \dfrac{v \sin( \theta) }{10} = 5$

$\implies \: v \sin( \theta) = 50 \: \: \: \: ....(2)$

Now, let max height be H:

$\therefore \: H = \dfrac{ {v}^{2} { \sin}^{2}( \theta) }{2g}$

$\implies \: H = \dfrac{ \bigg \{ {v \sin( \theta) \bigg \} }^{2} }{2g}$

$\implies \: H = \dfrac{ \bigg \{{50 \bigg \} }^{2} }{2g}$

$\implies \: H = \dfrac{ 2500 }{2g}$

$\implies \: H = \dfrac{ 2500 }{2 \times 10}$

$\implies \: H = 125 \: m$

So, max height attained is 125 metres.

Answer 2

Given:

Time of flight of a projectile = 10s

Range of projectile = 400m

To find:

The maximum height attained by the projectile

Solution:

As we know that, range = v² sin(2α)/g

where v = initial velocity

α = Angle of projection

g = acceleration due to gravity

v² sin(2α)/g = 400

Since time of the projectile = 400m

and time = 2v sin(α)/g = 10

v sin(α)/g = 5

Putting g = 9.8, we get:

v sin(α) = 49

Now, let the maximum height attained be H

since, H = v² sin²(α)/ 2g

H = [v sin(α)]² / 2g

H = 49² / 2g

H = 122.5 m

Therefore, the max height attained will be 122.5 m.