A particle is projected up with an initial velocity of $80 ext{ ft/sec}$. The ball will be at a height of $96

Question

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

Accepted Answer

1.  **Identify Given Values:**
    *   Initial velocity, $u = 80	ext{ ft/s}$ (upwards).
    *   Displacement, $s = h = 96	ext{ ft}$.
    *   Acceleration, $a = -g = -32	ext{ ft/s}^2$ (downwards).
2.  **Apply Kinematic Equation:** Use the second equation of motion, $s = ut + \frac{1}{2}at^2$ .
    *   $96 = 80t + \frac{1}{2}(-32)t^2$
    *   $96 = 80t - 16t^2$
3.  **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: $t = 2	ext{ sec}$ and $t = 3	ext{ sec}$.
4.  **Conclusion:** The particle is at the height of $96	ext{ ft}$ twice: once while going up (at 2s) and once while coming down (at 3s).

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