Answer:
[tex]1.29 \ s, 2.05 \ m \ {and} \ 17.58 \ m.[/tex]
Explanation:
Initial velocity of object, [tex]v=15 \ m/s.[/tex]
Angle from horizontal, [tex]\theta=25^o.[/tex]
Now , formula of time of flight, [tex]t = \dfrac{2\times v \times sin\theta}{g}[/tex]
Putting all these values in above equation.
[tex]t=\dfrac{2 \times 15 \times sin25^o}{9.8 } \ s=1.29 \ s[/tex]
Also, maximum height , [tex]H_{max}= \dfrac{v^2\times sin^2\theta}{2\times g}.[/tex]
Putting all required values,
[tex]H_{max}= \dfrac{15^2\times sin^225}{2\times 9.8}=2.05 \ m[/tex]
Again, range is given by , [tex]R=\dfrac{v^2\times sin2\theta}{g}[/tex]
Putting required values we get,
[tex]R=\dfrac{15^2\times sin(2\times 25)}{9.8}=17.58 \ m[/tex]
Hence, this is the required solution.
