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:

Let speed at half of max height be V then:

V

2

=50

2

−2g

2

125

⇒V

2

=2500−1250=1250

⇒V=

1250

=35.35m/s.

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,

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

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