Question

a ball thrown upwards reaches a point p of its path at end of 4 seconds and highest point q at the end of 12 seconds . after how many seconds from the start will it reach point p again?​

Answer 1

Answer:

Explanation:

The basic equation of projectile motion is h(t)=g/2 t²+ vt + c, where h(t) is the height after t seconds, g is gravity, v is initial velocity, and c is the starting height. Max altitude is reached at t= -b/2a from the quadratic equation ax²+bx+c. So:

at t=12 we have 12=-b/2a=-b/-9.8

-b=12*-9.8=-117.6

b=117.6

So, the initial velocity of the ball is 117.6 m/s

Now h(4)=-4.9(4²)+117.6(4)=470.4–78.4=392  

Then:

392=-4.9t²+117.6t

=>4.9t²-117.6t+392=0

=>49t²-1176t+3920=0

=>(t-28)(7t-140)=0

=>t=20

Since the ball reaches its’ highest point after 12 seconds, then the total time in the air is 2 x 12, or 24 seconds; and the height it reaches after 4 seconds is the same as the height it is at 4 seconds before landing, or 20 seconds...

(OR)

Firstly it goes above then at that reference of motion it takes 12 seconds.............

But the total height of the motion of that particle is 720 metres and point P is located at 400 meters from the lowest point possible.

So, 720-400 = 320m..

Now . 320 = 5t^2

64 = t^2 =8, So,  the total time= 12+8= 20 seconds.........

Answer 2

Answer:

The ball will reach at point 'p' again from start after 20 seconds.

Explanation:

Initially ball was at ground time, t = 0s. when the ball thrown upward, it reaches at point 'p' at time t=4s and the further moving upward it reached highest point 'q' at t=12s.

The velocity of ball at highest point, v=0 because the ball is moving upward against the gravity cause retardation and a = -g,

$v = u + at\\0 = u -g(12)\\u=12g$                                                           ..................(1)

Consider that v₄ is the velocity of ball at t=4s then,

$v_{4}= u-gt$

$v_{4}=12g-4g$

$v_{4}=8g$

Consider that Δt is time to reach from point 'p' to again point 'p'.

$S= ut+\frac{1}{2}gt^2$

$S= v_{4}(\triangle t)-\frac{1}{2}g(\triangle t)^2$

$0=(8g)(\triangle t)-\frac{g}{2}(\triangle t)^2$

$(8g)(\triangle t)=\frac{g}{2}(\triangle t)^2$

Δt = 16s

Time to reach from point 'p' to again point 'p' = 16s

Total time from the start to reach point p again = 16s + 4s = 20s

Therefore, total time taken by ball to reach from starting point to point 'p' again is equal to 20s.