91Ó°ÊÓ

The initial height of the grasshopper is$y_0=0$ .

The initial horizontal position of the grasshopper is$x_0=0$ .

The initial velocity makes an angle${\mathrm{Î\pm }}_0=50.0°$ with the horizontal.

The vertical component of velocity at the top is$v_y=0$ .

If a particle is thrown in projectile motion, then the velocity of the particle will contain two components, first one is horizontal and other one is vertical.

The second law of motion is given by,

$s=ut+\frac{1}{2}a{t}^2$

Here s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

The third law of motion is given by,

$v^2=u^2+2as$

Here v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

(a)

Using the third law of motion,

$v^2=u^2+2as$

Substitute$0m/s$ for$v,-9.8m/s^2$ , for a and$0.0674m$ for s in the above equation,

$$\begin{gathered} 0^2=u^2+2×-9.8×\left(0.0674\right) \\ u=1.15m/s \end{gathered}$$

For the vertical component,

$u=u_{vertical}\sin \mathrm{θ}$

Substitute 1.15 m/s for u and$50°$ for$\mathrm{θ}$ in the above equation.

$$\begin{gathered} u_{vertical}=\frac{1.15}{\sin 50°} \\ {u}_{vertical}=1.50m/s \end{gathered}$$

Therefore the initial speed of the grasshopper is 1.50 m/s .

• (b)

Apply the second law of motion in horizontal direction,

$s=u\cos \mathrm{θ}t+\frac{1}{2}a{t}^2$

Substitute 1.06 m for s , 1.50 m/s for u , $50°$for $\mathrm{θ}$ and $0m/s^2$for a in the above equation.

$$\begin{gathered} 1.06=1.50×\cos 50°×t+\frac{1}{2}×0×{\left(t\right)}^2 \\ t=1.10s \end{gathered}$$

Apply the second law of motion in vertical direction,

$s=u\cos \mathrm{θ}t+\frac{1}{2}a{t}^2$

Substitute$1.15m/s$ for u ,$50°$ for$\mathrm{θ}$ , 1.10 s for t and$-9.8m/s^2$ for a in the above equation,

$$\begin{gathered} s=1.15×1.10+\frac{1}{2}×-9.8×{\left(1.10\right)}^2 \\ s=-4.66m \end{gathered}$$

Therefore the height of the cliff relative to the ground is 4.66m .