Question

A ball is thrown vertically upwards reaches a height of 80m
calculate the time to reach highest point and
the speed of the ball upon arrival on the ground.

Answer 1

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

$\green{\tt{\therefore{Initial\:velocity=40\:m/s}}}$

$\green{\tt{\therefore{Time\:taken=4\:sec}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Height (s) = 80 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt :\implies Time \: taken(t) = ? \\ \\ \tt: \implies Initial \: velocity(u) =?$

• According to given question :

$\tt \circ \: Acceleration = - 10 { \: m/s}^{2} \\ \\ \tt \circ \: Final \: velocity = 0 { \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {u}^{2} + 2\times ( - 10) \times 80 \\ \\ \tt: \implies - {u}^{2} = - 1600 \\ \\ \tt: \implies {u}^{2} = 1600 \\ \\ \tt: \implies u = \sqrt{1600} \\ \\ \green{\tt: \implies u = 40 \: m/s} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = 40 + ( - 10) \times t \\ \\ \tt: \implies - 40 = - 10 \times t \\ \\ \tt: \implies t = \frac{ - 40}{ - 10} \\ \\ \green{\tt: \implies t = 4 \: sec}$

Answer 2

$\sf{\red{Solution:}}$

When the ball is thrown vertically upward ; then the maximum height attained by the ball is

$\tt \pink h = \frac{ {u}^{2} }{2}\:g \\ \\ \tt \purple{h = 80 \: m} \\ \\ \tt \pink{80 = \frac{ {u}^{2} }{2} \times 10} \\ \\ \tt \orange{ {u}^{2} = 80 \times 2 \times 10 = 1600 \: m \: per \: sec} \\ \\ \tt \red{u = 40 \: m \: per \: sec} \\ \\ \tt \blue{Time \: of \: ascent = \: \frac{u}{g} }\\ \\ \tt \purple{ t= \frac{40}{10}} \\ \\ \tt \green {\boxed{4 \: sec}}$