A ball is thrown straight upward with a velocity v0 from a height h above the ground. The time taken for the ball to strike the ground is

Question

A ball is thrown straight upward with a velocity v0 from a height h above the ground. The time taken for the ball to strike the ground is (A) (v0/g)(1+(2 g h/v02)) (B) (v0/g)(√1+(2 g h/v0 2)) (C) (v0/g)(1-√1+(2 g h/v02)) (D) (v0/g)(1+√1+(2 g h/v02))

Tardigrade Admin · Accepted Answer

Apply s=u t+(1/2) a t2 ⇒ -h=v0 t-(1/2) g t2 ⇒ g t2-2 v0 t-2 h=0 ⇒ t=(2 v0 ± √4 v02+8 g h/2 g) As t is not negative, so ignoring negative sign, we get t=(2 v0+2 √v02+2 g h/2 g) =(v0/g)[1+√1+(2 g h/v02)]

Q. A ball is thrown straight upward with a velocity $v_{0}$ from a height $h$ above the ground. The time taken for the ball to strike the ground is

$\frac{v _{0}}{g} (1 + \frac{2 g h}{v _{0}^{2}})$

$\frac{v _{0}}{g} (1 + \frac{2 g h}{v _{0} ^{2}})$

$\frac{v _{0}}{g} (1 - 1 + \frac{2 g h}{v _{0}^{2}})$

$\frac{v _{0}}{g} (1 + 1 + \frac{2 g h}{v _{0}^{2}})$

Q. A ball is thrown straight upward with a velocity v0​ from a height h above the ground. The time taken for the ball to strike the ground is

gv0​​(1+v02​2gh​)

gv0​​(1+v0​22gh​​)

gv0​​(1−1+v02​2gh​​)

gv0​​(1+1+v02​2gh​​)

Apply s=ut+21​at2 ⇒−h=v0​t−21​gt2 ⇒gt2−2v0​t−2h=0 ⇒t=2g2v0​±4v02​+8gh​​ As t is not negative, so ignoring negative sign, we get t=2g2v0​+2v02​+2gh​​ =gv0​​[1+1+v02​2gh​​]

Q. A ball is thrown straight upward with a velocity $v_{0}$ from a height $h$ above the ground. The time taken for the ball to strike the ground is

$\frac{v _{0}}{g} (1 + \frac{2 g h}{v _{0}^{2}})$

$\frac{v _{0}}{g} (1 + \frac{2 g h}{v _{0} ^{2}})$

$\frac{v _{0}}{g} (1 - 1 + \frac{2 g h}{v _{0}^{2}})$

$\frac{v _{0}}{g} (1 + 1 + \frac{2 g h}{v _{0}^{2}})$