A projectile is thrown at angle β with vertical. It reaches a maximum height H . The time taken to reach the highest point of its path is

Question

A projectile is thrown at angle β with vertical. It reaches a maximum height H . The time taken to reach the highest point of its path is (A) √(H/g) (B) √(2 H/g) (C) √(H/2 g) (D) √(2 H/g cos β)

Tardigrade Admin · Accepted Answer

t=(u sin (90-β)/g)=(u cos β/g) H=(u2 sin 2(90-β)/2 g)=((u cos β)2/2 g) ⇒ u cos β=√2 g H ⇒ t=(√2 g H/g)=√(2H/g)

Q. A projectile is thrown at angle $β$ with vertical. It reaches a maximum height $H .$ The time taken to reach the highest point of its path is

$\frac{H}{g}$

$\frac{2 H}{g}$

$\frac{H}{2 g}$

$\frac{2 H}{g c o s β}$

$t = \frac{u s i n ( 90 - β )}{g} = \frac{u c o s β}{g}$
$H = \frac{u ^{2} s i n ^{2} ( 90 - β )}{2 g} = \frac{( u c o s β ) ^{2}}{2 g}$
$\Rightarrow u cos β = 2 g H$
$\Rightarrow t = \frac{2 g H}{g} = \frac{2 H}{g}$

Q. A projectile is thrown at angle β with vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is

gH​​

g2H​​

2gH​​

gcosβ2H​​