Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A particle is projected up with an initial velocity of $80\text{ ft/sec}$ . The ball will be at a height of $96\text{ ft}$ from the ground after (Given $g=32\text{ ft/s}^2$ ):

2.0 and 3.0 sec

Only at 3.0 sec

Only at 2.0 sec

After 1 and 2 sec

Step-by-Step Solution

Identify Given Values: Initial velocity, $u = 80\text{ ft/s}$ (upwards). Displacement, $s = h = 96\text{ ft}$ .

Acceleration, $a = -g = -32\text{ ft/s}^2$ (downwards).

Apply Kinematic Equation: Use the second equation of motion, $s = ut + \frac{1}{2}at^2$ . $96 = 80t + \frac{1}{2}(-32)t^2$ $96 = 80t - 16t^2$
Solve Quadratic Equation: Divide the entire equation by 16: $6 = 5t - t^2$ . Rearrange: $t^2 - 5t + 6 = 0$ . Factorize: $(t - 3)(t - 2) = 0$ . Roots: $t = 2\text{ sec}$ and $t = 3\text{ sec}$ .
Conclusion: The particle is at the height of $96\text{ ft}$ twice: once while going up (at 2s) and once while coming down (at 3s).

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