Question

A ball is gently dropped from a height of 50m. If its acceleration increases at the rate of 9.8ms-2, with what velocity will it strike the ground

Answer 1

Given :

• Height,h = 50 m
• Acceleration, a = 9.8 m/s²
• Initial velocity,u = 0 m/s

To Find ;

The velocity will it strike the ground?

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

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

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

We have to Find the velocity by which the ball will strike the ground.

From the third equation of motion,we have:

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

Putting the values in the above formula,we get:

$\sf\:v{}^{2}={0}^{2}+2\times9.8\times50$

$\sf\:v{}^{2}=0+2\times\dfrac{98}{10}\times50$

$\sf\:v{}^{2}=0+2\times98\times5$

$\sf\:v{}^{2}=10\times98$

$\sf\:v{}^{2}=980$

$\sf\:v=\sqrt{980}$

$\sf\:v=31.30ms{}^{-1}(approx)$

Thus,the velocity of a ball is 31.30 m/s

Answer 2

Answer:-

Given:-

• Initial velocity (u) = 0 m/s [∵ It was stationary]
• Height (h) = 50 m
• Acceleration due to gravity = + 10 m/s² [Downward]

To Find: Final velocity when it'll strike the Earth.

We know,

2as = v² - u²

where,

• a = Acceleration,
• s = Distance travelled,
• v = Final velocity,
• u = Initial one.

Also,

v = √(2as + u²)

= √2as (In this case)

= √[2(9.8 m/s²)(50 m)]

= √(19.6 m/s² × 50 m)

= √980 m²/s²

= 31.304 m/s (Approx.)

∴ The final velocity when the ball will strike the Earth would be approximately 31 m/s.