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.

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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
$\begin{array}{l} h = - ut + \frac{1}{2} {gt}^{2} \\ \Rightarrow 25 = - 20 t + \frac{1}{2} 10 t^{2} \\ \Rightarrow 25 = - 20 t + 5 t^{2} \\ \Rightarrow 5 t^{2} - 20 t - 25 = 0 \\ \Rightarrow t^{2} - 4 t - 5 = 0 \\ \Rightarrow t^{2} - 5 t + t - 5 = 0 \\ \Rightarrow t (t - 5) + 1 (t - 5) = 0 \\ \Rightarrow (t + 1) (t - 5) = 0 \\ \Rightarrow t = - 1, 5 \sec \\ ∴ t = 5 \sec \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE