Question

A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10m s-2 ,with what velocity will it strike the ground ? After what time will it strike the ground ?

Answer 1

given that,

a of the ball = 10m/s^2

u of the ball = 0m/s

distance = 20m

velocity with which it strikes ground =

v^2 = u^2 + 2as

= (0 + 2 * 10 * 20) m/s

= √400m/s = 20m/s

time taken by the ball to strike ground=

v = u + at

20m/s = 0 + 10m/s^2 * t

t = 2s

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Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★Given :-

u = 0 m/s (Initial Velocity)

a = 10 m/s² (Acceleration)

s = 20 m (Height)

★To Find :-

v = ? (Final Velocity)

★Solution :-

$\textbf{\boxed{Third\; Equation\;of\;Motion}}$

${\boxed{v^{2} = u^{2} + 2as}}$

${\boxed{v^{2} - 0^{2} + 2\times 10\times 20m}}$

v² = 400 m²

v = √400 m²

v = 20 m/s

$\textbf{\boxed{First\; Equation\;of\;Motion}}$

${\boxed{v= u+ at}}$

20 = 0 + 10 × t

20 = 10 × t

$\tt{\rightarrow\dfrac{20}{10} = t}$

t = 2 second

$\boxed{\begin{minipage}{11 cm} Additional Information \\ \\ \ Final\;Velocity=Initial\; velocity+Acceleration\times Time \\ \\ Momentum = Mass\times Velocity \\ \\ Force=Mass\times Acceleration \\ \\ Force = \dfrac{Change\;in\;Momentum}{Time\;Interval} \\ \\ Impulse = Force\times Time \end{minipage}}$