A ball is thrown vertically upwards with a velocity of 20m/sec from the top of multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground.
a) How high will the ball rise ? and
b) How long will it be in air before the ball hit’s the ground ? Take g = 10ms^-2.

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Detailed Solution

$a) u = 20 {ms}^{- 1}, g = 10 {ms}^{- 2}$

from the surface of projection,
$h_{max} = \frac{u^{2}}{2 g} = \frac{(20)^{2}}{2 (10)} = \frac{20 \times 20}{20} = 20 m$

From the ground,

$H_{max} = 20 + 25 = 45 m$
b) h = 25m, u = 20ms^-1, g = 10m
$\begin{array}{l} h = - ut + \frac{1}{2} {gt}^{2} \\ \Rightarrow 25 = - 20 t + \frac{1}{2} 10 t^{2} \\ \Rightarrow 25 = - 20 t + 5 t^{2} \\ \Rightarrow 5 t^{2} - 20 t - 25 = 0 \\ \Rightarrow t^{2} - 4 t - 5 = 0 \\ \Rightarrow t^{2} - 5 t + t - 5 = 0 \\ \Rightarrow t (t - 5) + 1 (t - 5) = 0 \\ \Rightarrow (t + 1) (t - 5) = 0 \\ \Rightarrow t = - 1, 5 \sec \\ ∴ t = 5 \sec \end{array}$