A ball is dropped from the top of a tower of height 100 m and at the ...
Given:
Height of the tower, h = 100 m
Initial velocity of the ball projected upwards, u = 25 m/s
Acceleration due to gravity, g = 9.8 m/s²
To find:
Distance from the top of the tower, at which the two balls meet
Explanation:
Let's first calculate the time taken by each ball to reach the meeting point.
For the ball dropped from the top of the tower:
Using the equation of motion,
h = ut + 1/2 gt²
where h = 100 m, u = 0 m/s (as it is dropped), g = 9.8 m/s²
On substituting the values, we get
100 = 0 × t + 1/2 × 9.8 × t²
Simplifying, we get
t² = 100/4.9
t = √(100/4.9) = 4.52 s
For the ball projected upwards from ground:
Using the equation of motion,
v = u + gt
where v = 0 m/s (at the highest point), u = 25 m/s, g = 9.8 m/s²
On substituting the values, we get
0 = 25 – 9.8 × t
t = 25/9.8 = 2.55 s
Now, let's calculate the distance from the top of the tower, at which the two balls meet.
Distance travelled by the ball dropped from the top of the tower in 4.52 s:
Using the equation of motion,
s = ut + 1/2 gt²
where u = 0 m/s, g = 9.8 m/s², t = 4.52 s
On substituting the values, we get
s = 0 × 4.52 + 1/2 × 9.8 × (4.52)² = 100 m
Distance travelled by the ball projected upwards from ground in 2.55 s:
Using the equation of motion,
s = ut + 1/2 gt²
where u = 25 m/s, g = 9.8 m/s², t = 2.55 s
On substituting the values, we get
s = 25 × 2.55 + 1/2 × 9.8 × (2.55)² = 78.4 m
Therefore, the distance from the top of the tower, at which the two balls meet is 78.4 m.
Hence, the correct answer is option D.
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