Question

A stone is dropped from the top of a tower 100m high and instantly a second stone is projected vertically upward from the bottom with a velocity of 25ms-1 find when and where the two stone meet . Take g=10ms-2

Answer 1

Hello Mate!

So the distance from top to ground is 100 m high.

Now, let the distance that the ball travelled from top to down be x.

s = ut + ½ at²

x = (0)(t) + ½ at²

x = ½ at² __(i)

Now, the distance that the ball thrown from down will travel 100 - x m.

100 - x = (25)(t) + ½ (-a)t²

100 - x = 25t - ½ at² __(ii)

Hence by adding both equation we get,

100 = 25t

4 seconds = time

Keeping value in equation (i)

x = ½ × ( 10 m/s² ) × ( 4 )²

x = 80 m from the top.

Have great future ahead!

Answer 2

$<b>Heya mate. (^_-). Solution below.$
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♦ For first stone, the one dropped from the tower,


Let the distance covered (s) be x.

initial velocity (u) = 0 m/s

acceleration due to gravity (g) = + 10 m/s²


Then, by second equation of motion,

s = ut + ½gt² (t = time)

=> x = (0 × t) + ½ × 10 × t²

=> x = 0 + 5t²

=> x = 5t².. (eq i)



♦ For second stone,the one projected upwards,


s = total height of tower - x = 100 - x

u = 25 m/s

g = - 10 m/s²


By second equation of motion,

s = ut + ½at²

=> (100 - x) = (25 × t) + ½ × - 10 × t²

=> 100 - x = 25t - 5t²

=> - x = 25t - 5t² - 100

=> - (x) = - (-25t² + 5t² + 100)

Cancelling (-) sign,

=> x = - 25t² + 5t² + 100.. (eq ii)



♦ From i and ii,

x = x

=> 5t² = -25t + 5t² + 100

=> 5t² + 25t - 5t² = 100

Cancelling 5t² and - 5t²

=> 25t = 100

=> t = 100/25 = 4 s



•°• The time at they meet is 4 seconds.



♦ Now, putting the value of t in eq i,

=> x = 5t²

=> x = 5 (4)²

=> x = 5 × 16

=> x = 80 m


•°• The two stones meet after 4 seconds at 80 m from the tower top.
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Thank you.. ^_^