Solve the following :


A ball is thrown horizontally from a point $100 \mathrm{~m}$ above the ground with a speed of $20 \mathrm{~m} / \mathrm{s}$ Find

(a) The time it takes to reach the ground

(b) The horizontal distance it travels before reaching the ground

(c) The velocity (direction and magnitude) with which it strikes the ground.


$x$-axis $y$-axis

$\mathrm{u}_{\mathrm{x}}=20 \mathrm{~m} / \mathrm{s} \mathrm{u}_{\mathrm{y}}=0 \mathrm{~m} / \mathrm{s}$

$a_{x}=0 m / s^{2} a_{y}=g m / s^{2}$

(a) $s_{y}=u_{y} t+\frac{1}{2} a_{y} t^{2}$

$100=0+\frac{1}{2}(9.8) t^{2}$

$t=4.5 \mathrm{sec}$

(b) $s_{x}=u_{x} t$


$\mathrm{s}_{\mathrm{x}}=90 \mathrm{~m}$

(c) $v_{x}=u_{x}+a t v_{y}=u_{y}+a_{y} t$

$v_{x}=20 \quad v_{y}=0+(9.8)(4.5)$

$v_{y}=44.1 \mathrm{~m} / \mathrm{s}$

$\sqrt{v_{y}=44.1 \mathrm{~m} / \mathrm{s}}$

$\mathrm{v}=49 \mathrm{~m} / \mathrm{s}$ at $\tan ^{-1}\left(\frac{44.1}{20}\right)$ that is $66^{\circ}$ from ground

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now