Question

a stone is released from the top of a tower 100 m high at the same instant another stone is thrown up from the base is the tower with a speed of 25m/s. find when and where the two stones will meet in air (g = 10m/s²)​

Answer 1

$\huge{\underline{\boxed{\mathfrak{\blue{Solution:-}}}}}$

$\bigstar{\underline{\mathfrak{Given:-}}}$

• let "t" = time after which both stones meet

• "S" = distance travelled by the stone dropped from the top of tower

• (100-S) = distance travelled by the projected stone.

$\huge{\underline{\boxed{\mathfrak{\blue{Explaination:-}}}}}$

i) For stone dropped from the top of tower

-S = 0 + 1/2 (-10) t²

$\boxed{\sf{S = 5t^{2}}}$

ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t²

$\boxed{\sf{25t - 5t^{2}}}$

Adding i) and ii) , We get

100 = 25t

$\boxed{\sf{t = 4 s}}$

Therefore, Two stones will meet after 4 s.

¡¡¡) Put value of t = 4 s in Equation i) , we get

S = 5 × 16

$\boxed{\sf{= 80 m}}$

Thus , both the stone will meet at a distance of 80 m from the top of tower.

______________________________________________________________________________

Answer 2

Answer:

hey mate

Explanation:

here is your answer

solution :

let the stones meet at R at a distance X from the base of the tower PQ fig 3.15

let t be the time at which the stones meet.

Data : For the stone falling from rest u1 = 0

S1 = - (100 - X) for the stone thrown up

U2 = 25 m/s

S2 = X

s = t - 1/2 gt²

-(100 - X) = -1/2 gt²

= 100 - X = 1/2 gt² (eqn 1)

= S2 = u2t - 1/2 gt²

= X= 25t - 1/2 gt². (eqn 2)

Adding eqs (1) and (2)

100 = 25t

=t = 4s

substituting for t in eq (2)

X = (25m/s) (4s) - 1/2 (10m/s²) (4a)²

= 100 -5 × 16 = 100 - 80

= 20 m

Thus the stones will meet in air 4 second from the start at a height of 20 m from the ground

$thanku$