Question

A particle is projected from groundwith velocity V making an angletheta with the horizontal. if it justcrosses the top of the two poles ofequal height H after 2 seconds and6 seconds respectively then the maximum height attained by particle will be

answer please i Will make you brainlist. help me..

Answer 1

hey mate!

if at two instant t1 abd t2 ..

if particle is at same height , the height of particle at instant is always equal to

y = g/2 (t1t2)

y = 10/2(2*10)

y = 100m

hope it will help you

#Phoenix

Answer 2

Answer:

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

vertical component of velocity, u_y=vsin\thetauy=vsinθ

applying Formula,

s = ut + 1/2 at²

for vertical direction

y=u_yt+\frac{1}{2}a_yt^2y=uyt+21ayt2

here, a_y=-gay=−g

or, y =vsin\theta t-\frac{1}{2}gt^2y=vsinθt−21gt2

if we put t = 1sec then y = h

h=vsin\theta-\frac{1}{2}gh=vsinθ−21g ....(1)

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

so, h=3vsin\theta-\frac{9}{2}gh=3vsinθ−29g ....(2)

from equations (1) and (2),

2vsin\theta=4g2vsinθ=4g

now time of flight , T = \frac{2vsin\theta}{g}g2vsinθ

= 4g/g = 4sec

hence, answer is 4sec