Question

A ball is thrown vertically upwards with a velocity of 20ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0m from the ground.
(a) How high will the ball rise?
(b) how long will it be before the ball hits the ground?

Answer 1

$- h = ut - 0.5a{t}^{2}$
h=25 and u=20m/sec^-1
solving you will get t=5sec

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Maximum\:height=45\:m}}}$

$\green{\tt{\therefore{Total\:time\:taken=5\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}$

$\tt: \implies Initial \: velocity = 20 \: m/s$

$\tt: \implies Height \: of \: building = 25 \: m$

$\red{\underline \bold{To \: Find :}}$

$\tt: \implies Maximum \: height =?$

$\tt: \implies Time \: taken \: to \: reach \: ground =?$

• Given Question

$\tt \circ \: Final \: velocity = 0 \: m/s$

$\tt: \implies {v}^{2} = {u}^{2} + 2as$

$\tt: \implies {0}^{2} = {20}^{2} + 2 \times ( - 10) \times s$

$\tt: \implies - 400 = - 20 \times s$

$\tt: \implies s = \frac{ - 400}{ - 20}$

$\green{\tt: \implies s = 20 \: m}$

$\ \green{\tt \therefore Maximum \: height \:attained\: by \: ball \: is \: 45 \: m}$

$\tt: \implies v = u + at$

$\tt: \implies 0 = 20 + ( - 10) \times t_{1}$

$\green{\tt: \implies t_{1} = 2 \: sec}$

$\tt \circ \: Initial \: velocity = 0 \: m/s$

$\tt: \implies s_{total} = ut + \frac{1}{2} a {t}^{2}$

$\tt: \implies 45 = 0 \times t + \frac{1}{2} \times 10 \times { t_{2} }^{2}$

$\tt: \implies \frac{90}{10} = { t_{2}}^{2}$

$\green{\tt: \implies t_{2} = 3 \: sec}$

$\green{\tt \therefore Total \: time \: taken \: to \: reach \: ground \: is \: 5 \: sec}$