Question

A projectile is thrown with velocity v making an angle teta with the horizontal .it just crosses the top of two poles, each of height h, after 1 second and 3 second respectively .the time of flight of the projectile is

Answer 1

A projectile is thrown with velocity v making an angle $\theta$ with the horizontal.

vertical component of velocity, $u_y=vsin\theta$

applying Formula,

s = ut + 1/2 at²

for vertical direction

$y=u_yt+\frac{1}{2}a_yt^2$

here, $a_y=-g$

or, $y =vsin\theta t-\frac{1}{2}gt^2$

if we put t = 1sec then y = h

$h=vsin\theta-\frac{1}{2}g$....(1)

again, if we put t = 3sec then y = h

so, $h=3vsin\theta-\frac{9}{2}g$....(2)

from equations (1) and (2),

$2vsin\theta=4g$

now time of flight , T = $\frac{2vsin\theta}{g}$

= 4g/g = 4sec

hence, answer is 4sec

Answer 2

Answer:

Explanation:

the horizontal component of velocity is (v cosθ) and vertical component is (v sinθ)

Now,s=ut+12at2

a=−g

for  t=1 s ,g=10 m/s2,u=v sinθs=h

h=v sinθ−12×10×(1)2

h=v sinθ−5  ......(1)

similarly for t=3 s,h=3v sinθ−45 .....(2)

from (1) and (2),

v sinθ−5=3v sinθ−45

2v sinθ=40v sinθ=20 .....(3

)Now, maximum height,

H=v2sin2θ/2

H=400/20=20 m