Answer:
Explanation:
If u is the initial velocity at an angle \theta with horizontal then
Horizontal range of Frog can be given by
[tex]R=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
a=acceleration
t=time
Here initial horizontal velocity
[tex]u_x=u\cos \theta [/tex]
and there is no acceleration in the horizontal motion
Therefore
[tex]R=u\cos \theta \times t+0[/tex]
Considering vertical motion
[tex]Y=ut+\frac{1}{2}at^2[/tex]
here Initial vertical velocity [tex]u_y=u\sin \theta [/tex]
acceleration [tex]a=g[/tex]
for complete motion Y=0 i.e.displacement is zero
[tex]0=u\sin \theta \times t-\frac{1}{2}gt^2[/tex]
[tex]t=\frac{2u\sin \theta }{g}[/tex]
Therefore Range is
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
Range will be maximum when [tex]\theta =45[/tex]
[tex]R=\frac{u^2}{g}----1[/tex]
and Maximum height [tex]h_{max}=\frac{u^2\sin ^2 \theta }{2g}[/tex]
for [tex]\theta =45[/tex]
[tex]h_{max}=\frac{u^2}{4g}----2[/tex]
Divide 1 and 2
[tex]\frac{R_{max}}{h_{max}}=\frac{\frac{u^2}{g}}{\frac{u^2}{4g}}[/tex]
[tex]\frac{R_{max}}{h_{max}}=4[/tex]
