Answer:
Let speed at half of max height be V then:
V
2
=50
−2g
125
⇒V
=2500−1250=1250
⇒V=
1250
=35.35m/s.
[tex] \small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]
[tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex]
Initial velocity of ball (u)=50 m / s,
acceleration of ball.
= -g =9.8 m / s²
final velocity at the highest point (v)=0
[tex] \text{So applying the \( \rm 3^{r d} \) equation of motion we get:}[/tex]
[tex] \[ \begin{array}{l} \rm v^{2}=u^{2}-2 g s \\ \\ \rm v^{2}=u^{2}-2 g h_{\max } \\ \\ \rm \Rightarrow 0=50^{2}-2 \times 10 \times h_{\max } \\ \\ \rm\Rightarrow h_{\max }=\dfrac{2500}{20}=125 m \end{array} \][/tex]
Velocity for height,
[tex] \\ \rm\frac{h_{\max }}{2}=\frac{125}{2} [/tex]
[tex] \[ \begin{array}{l} \rm v^{2}=u^{2}+2 a s \\ \\ \rm =(50)^{2}+2(-10)\left(\frac{125}{2}\right) \\\\ \rm v^{2}=2500-1250=1250 \\\\ \rm v=\sqrt{1250} \: \: \: \: \: \boxed{\red{ \rm =35.4 m / s}} \end{array} \][/tex]
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Answers & Comments
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.
[tex] \small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]
[tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]
Initial velocity of ball (u)=50 m / s,
acceleration of ball.
= -g =9.8 m / s²
final velocity at the highest point (v)=0
[tex] \text{So applying the \( \rm 3^{r d} \) equation of motion we get:}[/tex]
[tex] \[ \begin{array}{l} \rm v^{2}=u^{2}-2 g s \\ \\ \rm v^{2}=u^{2}-2 g h_{\max } \\ \\ \rm \Rightarrow 0=50^{2}-2 \times 10 \times h_{\max } \\ \\ \rm\Rightarrow h_{\max }=\dfrac{2500}{20}=125 m \end{array} \][/tex]
Velocity for height,
[tex] \\ \rm\frac{h_{\max }}{2}=\frac{125}{2} [/tex]
[tex] \[ \begin{array}{l} \rm v^{2}=u^{2}+2 a s \\ \\ \rm =(50)^{2}+2(-10)\left(\frac{125}{2}\right) \\\\ \rm v^{2}=2500-1250=1250 \\\\ \rm v=\sqrt{1250} \: \: \: \: \: \boxed{\red{ \rm =35.4 m / s}} \end{array} \][/tex]