Question

A ball is projected vertically upwards with a speed of 50 m/s. Then the speed at half of the maximum height is

Answer 1

Answer:

Explanation:

Upward is Positive and Downward is Negative

Initial Speed u= 50m/s

g = -10 m/s²

At Maximum Height speed v = 0 m/s

using Third law of motion

v² = u² + 2gh

0 = (50 m/s)² + 2* (-10) (h)

20h = 2500

h=125m

So We have to find speed at half of maximum height :

h/2 = 125/2 = 62.25m

Again using third law of motion,

V = ?

u= 50m/s

H = 62.25m

g = -10 m/s²

V² = u² + 2gh

V² = (50)² + 2(-10)(62.25)

V² = 2500 -1250

V² = 1250

V = 35.35 m/s   Ans.

Answer 2

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Initial velocity of ball (u)=50 m / s,

acceleration of ball.

= -g =9.8 m / s²

final velocity at the highest point (v)=0

$\text{So applying the \( \rm 3^{r d} \) equation of motion we get:}$

$\[ \begin{array}{l} \rm v^{2}=u^{2}-2 g s \\ \\ \rm v^{2}=u^{2}-2 g h_{\max } \\ \\ \rm \Rightarrow 0=50^{2}-2 \times 10 \times h_{\max } \\ \\ \rm\Rightarrow h_{\max }=\dfrac{2500}{20}=125 m \end{array} \]$

Velocity for height,

$\\ \rm\frac{h_{\max }}{2}=\frac{125}{2}$

$\[ \begin{array}{l} \rm v^{2}=u^{2}+2 a s \\ \\ \rm =(50)^{2}+2(-10)\left(\frac{125}{2}\right) \\\\ \rm v^{2}=2500-1250=1250 \\\\ \rm v=\sqrt{1250} \: \: \: \: \: \boxed{\red{ \rm =35.4 m / s}} \end{array} \]$