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
Answers
Answered by
222
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!
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!
ShuchiRecites:
Thanks sis
Answered by
95
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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.
=========================================
Thank you.. ^_^
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