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

Question

If a ball is thrown vertically upwards with a speed u , the distance covered by it during the last t seconds of its ascent is (A) u t (B) (1/2) g t2 (C) u t - (1/2) g t2 (D) (. u + g t .) t

Tardigrade Admin · Accepted Answer

Let total height = H Total Time of ascent = T So, H=uT-(1/2)gT2 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-7w06y6u4vyvicuu6.jpg" /> Distance covered by ball in time (. T - t .) textsec is y = u (. T - t .) - (1/2) g (. T - t .)2 So, distance covered by ball in last t sec h = H - y = [u T - (1/2) g T2] - [u (. T - t .) - (1/2) g (. T - t .)2] By solving and putting T = (u/g) we will get, h = (1/2) g t2

Q. If a ball is thrown vertically upwards with a speed $u$ , the distance covered by it during the last $t$ seconds of its ascent is

$u t$

$\frac{1}{2} g t^{2}$

$u t - \frac{1}{2} g t^{2}$

$(u + g t) t$

Q. If a ball is thrown vertically upwards with a speed u , the distance covered by it during the last t seconds of its ascent is

ut

21​gt2

ut−21​gt2

(u+gt)t

Let total height =H Total Time of ascent =T So, H=uT−21​gT2 Distance covered by ball in time (T−t)sec is y=u(T−t)−21​g(T−t)2 So, distance covered by ball in last tsec h=H−y=[uT−21​gT2]−[u(T−t)−21​g(T−t)2] By solving and putting T=gu​ we will get, h=21​gt2

Q. If a ball is thrown vertically upwards with a speed $u$ , the distance covered by it during the last $t$ seconds of its ascent is

$u t$

$\frac{1}{2} g t^{2}$

$u t - \frac{1}{2} g t^{2}$

$(u + g t) t$