Question

a ball is thrown from the the top of a staircase which just touches the ceiling and finally hits the bottom of the steps . the initial speed of the ball is​

Answer 1

Answer:

if g =10m/s^2

then initial v =10m/s

Answer 2

Answer:

The initial speed of the ball when hits the bottom of the steps is 6.3 m /s

Explanation:

Given: A ball is thrown from the top of the staircase.

To find: The initial speed of ball when hits the bottom of the steps.

Explanation:

The ball touches the ceiling at the height of 5 m

Given that the distance between the floor and ball  is 4 m

4 - (u cos α)t      (1)

$2 = \frac{u^{2} sin^{2} \alpha }{2g}$          (2)

$- 3 = (usin\alpha )t - \frac{1}{2} gt^{2}$             (3)

From eq 2

$4g = u^{2} sin^{2} \alpha$

Taking square root on both sides we get,

2√g = u sinα

Sub these in eq 3 ,then we get,

$- 3 = 6.3t- 4.9t^{2}$

⇒ 4.9$t^{2}$ - 6.3t - 3 = 0

⇒ $t = \frac{6.3(+or-)\sqrt{39.7-4(4.9)(-3} }{2(4.9)}$

⇒ $t = \frac{6.3(+ or-)\sqrt{98.5} }{9.8}$

⇒ $t = \frac{6.3 (+ or -)9.9}{9.8}$

⇒ = 8.1 s

Sub value of t in eq 1

4 = (u cos α )(8.1 s)

⇒ u cos α = $\frac{4}{8}$ ≅ $\frac{1}{2}$

From eq 4 and 5, we ger

$u^{2} sin2^{2} \alpha +u^{2} sin^{2} \alpha = 40 + \frac{1}{4}$

⇒ u = √40.25

⇒ u = 6.3 m /s

Final answer:

The initial speed of the ball when hits the bottom of the steps is 6.3 m /s

SPJ3