Question

9. A ball is released from a height and it reaches the
ground in 3 s. If g = 9.8 m s?, find :
(a) the height from which the ball was released,
(b) the velocity with which the ball will strike the
ground​

Answer 1

Explanation:

We have,

(Consider upward motion as +ve direction and downward direction as -ve direction)

Initial speed of ball, $\sf{u = 0 \:ms^{-1}}$

Time taken, $\sf{t = 3 \:s}$

Acc due to gravity, $\sf{a=-g = -9.8 \:ms^{-2}}$

a.) Therefore,

Height from which the ball is released is given by,

$\sf{s = ut-\frac{1}{2}gt^2}$

$\implies{\sf{s = 0-\frac{1}{2}\times{(9.8)(3)^{2}}}}$

$\implies{\sf{s = -\times{(4.9)(9)}}}$

$\implies{\boxed{\sf{s = -44.1\: m\:}}\:✓✓✓}$

b.) The velocity with which the ball will strike the

ground is given by,

$\sf{v=u+at}$

$\implies{\sf{v=0-(9.8)(3)}}$

$\implies{\boxed{\sf{v=-29.4\:ms^{-1}}}\:✓✓✓}$

___________________________________________

HOPE IT HELPS ❣️❣️❣️