Physics - Kinematics Question with Solution | TestHub

A particle grazes the top of two poles of equal height at $t=2s$ and $t=6s$.

Distance of first pole from point of projection $=35m$.

Take $g=10m/s^2$.

### (P) Time of flight

For a projectile, times corresponding to the same height satisfy:

$$t_1+t_2=T$$

Therefore,

$$T=2+6=\boxed{8s}$$

$$\therefore \boxed{P\to (3)}$$

### (Q) Height of the pole

Vertical displacement at time $t$:

$$y=u_{y}t-\frac{1}{2}g{t}^2$$

At $t=2s$,

$$h=2u_y-20$$

At $t=6s$,

$$h=6u_y-180$$

Since both poles have the same height,

$$2u_y-20=6u_y-180$$

$$4u_y=160$$

$$u_y=40m/s$$

Hence,

$$h=2(40)-20=\boxed{60m}$$

$$\therefore \boxed{Q\to (4)}$$

### (R) Maximum height

$$H_{\max }=\frac{u_y^2}{2g}$$

$$H_{\max }=\frac{40^2}{2(10)}=\boxed{80m}$$

$$\therefore \boxed{R\to (1)}$$

### (S) Distance between the two poles

For the first pole,

$$x_1=u_x{t}_1$$

$$35=u_x(2)$$

$$u_x=17.5m/s$$

Time interval between the two poles:

$$\Delta t=6-2=4s$$

Therefore,

$$\Delta x=u_x\Delta t$$

$$\Delta x=17.5\times 4=\boxed{70m}$$

$$\therefore \boxed{S\to (2)}$$

### Final Answer:

$$\boxed{P-3,Q-4,R-1,S-2}$$