A ball thrown vertically upward returns to the thrower after 6 s. The ball is 5 m below the highest point at t = 2 s. The time at which the body will be at same position, (take g = 10 $m / s^{2}$ ).

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Solution

Total time of flight $(t) = 6 s$
So time to reach the maximum height $= 3 s$
if $u =$ projected velocity then:
$0 = u - g \frac{t}{2} \Rightarrow u = \frac{g t}{2} = \frac{60}{2} = 30$ m/s
Applying equation $v^{2} = u^{2} - 2 g h$ (for half journey)
$0 = 900 - 20 h$
$\Rightarrow h = 45 m$
When the ball is at height $5 m$ from highest point, it's height from ground $= 45 - 5 = 40$ m
Applying equation: $s = u t + \frac{1}{2} a t^{2}$ (at $h = 40$ m)
$\Rightarrow 40 = 30 t - \frac{1}{2} g t^{2}$
$\Rightarrow 40 = 30 t - 5 t^{2} \Rightarrow t^{2} - 6 t + 8 = 0$
As one root of the equation is given as: $t_{1} = 2 s$
By product of roots $t_{1} t_{2} = 8$
$\Rightarrow t_{2} = 4 s$ . (when the ball reaches the same height again)

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A ball thrown vertically upward returns to the thrower after 6 s. The ball is 5 m below the highest point at t = 2 s. The time at which the body will be at same position, (take g = 10m/s2).

A ball thrown vertically upward returns to the thrower after 6 s. The ball is 5 m below the highest point at t = 2 s. The time at which the body will be at same position, (take g = 10 $m / s^{2}$ ).