A body is projected at an angle of 60^{circ} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the initial velocity be $v$ and the angle of projection be $	heta = 60^{\circ}$.Given,the vertical component of initial velocity $v_y = v \si

Vedclass · Accepted Answer

$30.54$

Vedclass · Answer

(C) Let the initial velocity be $v$ and the angle of projection be $	heta = 60^{\circ}$.Given,the vertical component of initial velocity $v_y = v \sin 	heta = 40 \ m \ s^{-1}$.The time of flight $T = \frac{2 v \sin 	heta}{g} = \frac{2 	imes 40}{10} = 8 \ s$.The time at which we need to find the velocity is $t = \frac{T}{4} = \frac{8}{4} = 2 \ s$.The horizontal component of velocity remains constant throughout the motion: $v_x = v \cos 	heta$.Since $v \sin 	heta = 40$,we have $v = \frac{40}{\sin 60^{\circ}} = \frac{40}{\sqrt{3}/2} = \frac{80}{\sqrt{3}} \approx 46.19 \ m \ s^{-1}$.Thus,$v_x = \frac{80}{\sqrt{3}} 	imes \cos 60^{\circ} = \frac{80}{\sqrt{3}} 	imes \frac{1}{2} = \frac{40}{\sqrt{3}} \approx 23.09 \ m \ s^{-1}$.The vertical component of velocity at time $t$ is $v_y(t) = v \sin 	heta - g t = 40 - (10 	imes 2) = 20 \ m \ s^{-1}$.The magnitude of velocity at time $t$ is $v_t = \sqrt{v_x^2 + v_y(t)^2} = \sqrt{(23.09)^2 + (20)^2} = \sqrt{533.15 + 400} = \sqrt{933.15} \approx 30.54 \ m \ s^{-1}$.

Similar Questions

$A$ gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is $6.4 \text{ m}$. The speed of the bullets from the gun is . . . . . . m/s. (take $g = 10 \text{ m/s}^2$)

$A$ projectile is thrown with an initial velocity $(a \hat{i} + b \hat{j}) \ m/s$,where $\hat{i}$ and $\hat{j}$ are unit vectors along horizontal and vertical directions respectively. If the range of the projectile is twice the maximum height reached by it,then

The maximum horizontal range of a projectile is $16 \ km$. When the projectile is thrown at an elevation of $30^o$ from the horizontal,it will reach a maximum height of ....... $km$.

An aeroplane is flying horizontally with a velocity of $600 \, km/h$ at a height of $1960 \, m$. When it is vertically above a point $A$ on the ground,a bomb is released from it. The bomb strikes the ground at point $B$. The distance $AB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module