Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A man drops a ball downwards from the roof of a tower of height $400 \text{ m}$ . At the same time, another ball is thrown upwards with a velocity $50 \text{ m/s}$ from the surface of the tower. They will meet at which height from the surface of the tower?

100 m

320 m

80 m

240 m

Step-by-Step Solution

Analyze Relative Motion: Both balls are under the influence of the same acceleration due to gravity ( $g$ ) acting downwards. Therefore, the relative acceleration between them is zero ( $a_{rel} = g - g = 0$ ).
Calculate Time to Meet ( $t$ ): The initial relative velocity is the sum of their speeds (since they move towards each other): $u_{rel} = 50 + 0 = 50 \text{ m/s}$ . The relative distance to cover is the tower height $H = 400 \text{ m}$ . $t = \frac{\text{Relative Distance}}{\text{Relative Velocity}} = \frac{400}{50} = 8 \text{ s}$ .
Calculate Height from Surface ( $h$ ): Use the second equation of motion for the ball thrown upwards ( $u = 50 \text{ m/s}$ , $a = -g = -10 \text{ m/s}^2$ ). $h = ut - \frac{1}{2}gt^2$ $h = 50(8) - \frac{1}{2}(10)(8)^2$ $h = 400 - 5(64) = 400 - 320 = 80 \text{ m}$ Alternatively, calculate the distance fallen by the top ball ( $h' = \frac{1}{2}gt^2 = 320 \text{ m}$ ) and subtract from total height: $400 - 320 = 80 \text{ m}$ .

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