A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be

Question

A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be (A) g(t-T)2 (B) H-(1/2) g(t-T)2 (C) (1/2) g(t-T)2 (D) H-g(t-T)

Tardigrade Admin · Accepted Answer

Using, v=u+a t, we get 0=u-g T or u=g T Further, v2=u2+2 a s or 0=u2-2 g H or H=(u2/2 g)=(g2 T2/2 g)=(g T2/2) Let h be the distance travelled in time t. Then h=u t-(1/2) g t2=g T × t-(1/2) g t2 Subtracting (i) from (ii), we get h-H=g t T-(1/2) g t2-(g T2/2)=-(g/2)(t-T)2 ∴ h=H-(g/2)(t-T)2

Q. A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be

$g (t - T)^{2}$

$H - \frac{1}{2} g (t - T)^{2}$

$\frac{1}{2} g (t - T)^{2}$

$H - g (t - T)$

Q. A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be

g(t−T)2

H−21​g(t−T)2

21​g(t−T)2

H−g(t−T)

Using, v=u+at, we get 0=u−gT or u=gT Further, v2=u2+2as or 0=u2−2gH or H=2gu2​=2gg2T2​=2gT2​ Let h be the distance travelled in time t. Then h=ut−21​gt2=gT×t−21​gt2 Subtracting (i) from (ii), we get h−H=gtT−21​gt2−2gT2​=−2g​(t−T)2 ∴h=H−2g​(t−T)2

$g (t - T)^{2}$

$H - \frac{1}{2} g (t - T)^{2}$

$\frac{1}{2} g (t - T)^{2}$

$H - g (t - T)$