Question

If a jumping frog can give itself the same initial speed regardless of the direction in which it jumps (forward or straight up), how is the maximum vertical height to which it can jump related to its maximum horizontal range Rmax = v20/g?

Answer 1

Answer:

Explanation:

If u is the initial velocity at an angle \theta with horizontal then

Horizontal range of Frog can be given by

$R=ut+\frac{1}{2}at^2$

where u=initial velocity

a=acceleration

t=time

Here initial horizontal velocity

$u_x=u\cos \theta$

and there is no acceleration in the horizontal motion

Therefore

$R=u\cos \theta \times t+0$

Considering vertical motion

$Y=ut+\frac{1}{2}at^2$

here Initial vertical velocity $u_y=u\sin \theta$

acceleration $a=g$

for complete motion Y=0 i.e.displacement is zero

$0=u\sin \theta \times t-\frac{1}{2}gt^2$

$t=\frac{2u\sin \theta }{g}$

Therefore Range is

$R=\frac{u^2\sin 2\theta }{g}$

Range will be maximum when $\theta =45$

$R=\frac{u^2}{g}----1$

and Maximum height $h_{max}=\frac{u^2\sin ^2 \theta }{2g}$

for $\theta =45$

$h_{max}=\frac{u^2}{4g}----2$

Divide 1 and 2

$\frac{R_{max}}{h_{max}}=\frac{\frac{u^2}{g}}{\frac{u^2}{4g}}$

$\frac{R_{max}}{h_{max}}=4$