Answer:
The initial velocity is 50 m/s.
(C) is correct option.
Explanation:
Given that,
Time = 10 sec
For first half,
We need to calculate the height
Using equation of motion
[tex]v^2=u^2+2gh[/tex]
[tex]h =\dfrac{v^2}{2g}[/tex]....(I)
For second half,
We need to calculate the time
Using equation of motion
[tex]h =ut+\dfrac{1}{2}gt_{2}^2[/tex]
[tex]h=0+\dfrac{1}{2}gt_{2}^2[/tex]
[tex]t_{2}=\sqrt{\dfrac{2h}{g}}[/tex]
Put the value of h from equation (I)
[tex]t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}[/tex]
[tex]t_{2}=\dfrac{v}{g}[/tex]
According to question,
[tex]t_{1}+t_{2}=10[/tex]
[tex]t_{1}=t_{2}[/tex]
Put the value of t₁ and t₂
[tex]\dfrac{v}{g}+\dfrac{v}{g}=10[/tex]
[tex]\dfrac{2v}{g}=10[/tex]
[tex]v=\dfrac{10\times g}{2}[/tex]
Here, g = 10
The initial velocity is
[tex]v=\dfrac{10\times10}{2}[/tex]
[tex]v=50\ m/s[/tex]
Hence, The initial velocity is 50 m/s.
