Question

A ball is thrown up vertically and returns to a thrower after 6sec. find velocity with which it was thrown up,,maximum height it reaches and it s position after 4 sec..please answer fast......​.

Answer 1

Explanation:

As we know

Time of Flight=2u/g

Where G=Gravitational Accerlation

U=Intial Velocity

Now We know Time of Ascent=Time of Discent =u/g

Now given Time of FLIGHT=6 seconds

2u/g=6

u/g=3

Time of Ascent=3 seconds

Now it reach Maximum height in 3 seconds.

Now it reach Maximum height in 3 seconds.Final Velocity at maximum height=0 m/s

Now it reach Maximum height in 3 seconds.Final Velocity at maximum height=0 m/sAccerlation Due of Gravity=-9.8 m/s²

Negative Accerlation means At some point while going Upward its Velocity will Decrease and become zero.

We have to find Intial Velocity.

V=u+at

0=u+(-9.8)×3

u=29.4 m/s

So Intial Velocity will be=29.4 m/s

Velocity at which it was thrown=29.4 m/s

Now Maximum height it reach

V²=U²+2as

0²=(29.4)²+2×-9.8×s

0=864.36+(-19.6s)

-864.36=-19.6s

s=-864.36/-19.6

s=44.1 m

Maximum Height it reach=44.1 m

In first 3 seconds it travel=44.1 m

So In next 1 seconds.

Intial Velocity =0

(As at 3 seconds it reach Maximum Height).

Time=1 seconds

a=9.8 m/s² (as ball is coming downward in 4th second

S=ut+1/2at²

S=0×1+1/2×9.8×1²

S=4.9 m

So in 4th second it travels=4.9 m while Going Downward

Answer 2

$\boxed{\boxed{\huge{\red{Question :-}}}}$

A ball is thrown up vertically and returns to a thrower after 6sec. find velocity with which it was thrown up,,maximum height it reaches and it s position after 4 sec??

$\boxed{\boxed{\huge{\green{Answer:-}}}}$

S= 40 m

$\boxed{\boxed{\huge{\pink{Solution:-}}}}$

First a fall we have to find the initial velocity u using the conditions at highest point..

➡total time of flight =6 second

Therefore, time taken for the ball to reach the highest point is 3s. So-

v=u+gtv=u+gt Putting v=0,g=−10m/s2v=0,g=−10m/s2 & t=3s.t=3s.

⟹u=30m/s⟹u=30m/s

Now, the position of the ball at 4s4s will be same as in 2s2s due to symmetry in its trajectory. Therefore using the following eqn. we get the distance of the ball from the ground at 4s:-

⟹s=ut+(12)gt2⟹s=ut+(12)gt2

⟹s=30×2+(12)×(−10)22⟹s=30×2+(12)×(−10)22

⟹s=60−5×4⟹s=60−5×4

⟹s=40m.

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