Answer to Question #117297 in Mechanics | Relativity for Meryclare

Question

A particle is projected at an angle of 60O to the horizontal with a speed of 20 m/s and total time of pacicle flight at 3.46s Calculate of Speed of the particle at its maximum height [take g = 10 m/s2 ]

Answer 1

We have given,

Angle of projection $\theta=60^{\circ}$ ,initial speed $u=20m/s$ ,Time of flight $T_{flight}=3.46s$ .

We have to calculate the speed at maximum height $(H)$ .

Since, we know that height $H$ is achieved when the $y$ component of the final velocity becomes at time

$$t=\frac{1}{2}T_{flight}$$

But we noticed that $x$ component of the velocity of the particle does not change with time as there is no horizontal force acting on the particle, hence no acceleration i.e $a_x=0$

Thus, there is only horizontal component of velocity will survive, hence from the given data,

$$v_y=0$$

$$v_x=u\cos(\theta)-a_xt\\ \implies v_x=u\cos(\theta)\\ \implies v_x=20\cdot \cos(60^{\circ})=10ms^{-1}$$

Therefore, velocity of the particle at maximum height is $10ms^{-1}$