Question

A ball is thrown vertically upwards with a velocity of 20 ms -1 from the top of a multistoreyed buildings. The height of the point from where the ball is thrown is 25 m from the ground. (i) How high will the ball irse ? (ii) How long will it be before the ball hits the ground ? (iii) Trace the trajectory of this ball. Please answer this quickly if you want to be marked as the BRAINLIEST!

Answer 1

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

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

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

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: velocity = 20 \: m/s \\ \\ \tt: \implies Height \: of \: building = 25 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Maximum \: height =? \\ \\ \tt: \implies Time \: taken \: to \: reach \: ground =?$

• According to given question :

$\tt \circ \: Final \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {20}^{2} + 2 \times ( - 10) \times s \\ \\ \tt: \implies - 400 = - 20 \times s \\ \\ \tt: \implies s = \frac{ - 400}{ - 20} \\ \\ \green{\tt: \implies s = 20 \: m} \\ \\ \green{\tt \therefore Maximum \: height \:attained\: by \: ball \: is \: 45 \: m} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = 20 + ( - 10) \times t_{1}\\ \\ \tt: \implies - 20 = - 10 \times t_{1}\\ \\ \green{\tt: \implies t_{1} = 2 \: sec} \\ \\ \tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{total} = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt: \implies 45 = 0 \times t + \frac{1}{2} \times 10 \times { t_{2} }^{2} \\ \\ \tt: \implies \frac{90}{10} = { t_{2}}^{2} \\ \\ \green{\tt: \implies t_{2} = 3 \: sec} \\ \\ \green{\tt \therefore Total \: time \: taken \: to \: reach \: ground \: is \: 5 \: sec}$

Answer 2

$\bold {\green{answer}}\\\\ \bold {\orange{\therefore{ \blue{total \: \: time \: \: taken \: = \: 5sec}}}} \\\\ \blue{maximum \: \: height \: \: = \: 45 \: m} \\ \\ \large{\bold {\green{ \underline{ \underline{step - by - step - explanation}}}}} \\\\ \bold{\green{given}} \\\\ \implies{\text{initial \: velocity \: = 20ms}} \\\\ \implies{\text{height \: of \: building = 25m}} \\\\ \large{ \bold \pink{find}} \\\\ 1) \implies{maximum \: height} \\\\ 2)\implies{time \: taken \: to \: reach \: the \: ball \: hits \: the \: ground} \\\\ 3)\implies{trace \: \: the \: \: trajectory \: \: of \: \: ball} \\\\ \implies \bold{according \: \: to \: \: question} \\\\ \implies \bold \pink{final \: \: velocity \: = 0} \\\\ \implies \bold{time \: \: to \: \: reached \: \: maximum \: \: height \: \: obtained \: \: from} \\\\ \implies \bold \green{ \boxed{v \: = u \: + at}} \\\\ \implies \bold{ \boxed{a \: = - g}} \\\\ \implies \bold{0 \: =20 + ( - 10)(t) }$ $\implies{t = 2sec} \\\\ \implies{maximum \: \: height} \\\\ \implies \bold \green{ \boxed{s = ut + \frac{1}{2}a {t}^{2} }} \\\\ \implies{s \: = 20(2) + \frac{1}{2}( - 10) ({2})^{2} } \\\\ \implies{s \: = 40 - 20 = 20m} \\\\ \implies{total \: \: height \: = 20 + 25 = 45m} \\\\ \implies \bold \orange \therefore{total \: \: distance \: \: from \: \: maximum \: \: height \: = 45m} \\\\ \implies{s \: = ut \: + \frac{1}{2}at {}^{2} } \\\\ \implies{45 = 0 + \frac{1}{2}( - 10)(T1){}^{2} } \\\\ \implies{45 = 0 + ( - 5)(T1){}^{2} } \\ \implies{9 = (T1) {}^{2} } \\\\ \implies{T1 =3sec} \\\\ \implies \bold \therefore \green{total \: \: time \: \: = \: 2 + 3 \: = 5sec}$