First slide

Position Path Length (Distance) And Displacement

Question

If a ball is thrown vertically upwards with speed u, the distance covered during the last t second of its ascent is

Moderate

A

$\frac{1}{2} {gt}^{2}$
B

$ut - \frac{1}{2} {gt}^{2}$
C

(u+ gt)t
D

ut

Solution

The distance H covered by the ball in upward motion is given by
$H = uT - \frac{1}{2} {gT}^{2}$
where T is the time taken by the ball.
Further v = u-gT or 0=u-gT
or $T = \frac{u}{g} . . . (2)$
Therefore, $H = u \times (\frac{u}{g}) - \frac{1}{2} g {(\frac{u}{g})}^{2} . . . . . (3)$
Before the last t second, the time taken by the ball will be (T - t) second. Let the distance be x, Then
$\begin{array}{l} x = u (T - t) + \frac{1}{2} g (T - t)^{2} \\ or x = \frac{1}{2} g (T - t)^{2} = \frac{1}{2} g {(\frac{u}{g} - t^{2})}^{2} (∵ u = v = 0) \end{array}$
Required dista4ce = (H - x)
$\begin{array}{l} ∴ (H - x) = [\{\frac{u^{2}}{g} - \frac{1}{2} \frac{u^{2}}{g}\} - \{\frac{1}{2} g {(\frac{u}{g} - t)}^{2}\}] \\ = ut - \frac{1}{2} {gt}^{2} \end{array}$

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