A projectile is thrown from a point O on the | vedclass

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

Question

(A) Initial velocity $u = 5\sqrt{2} 	ext{ m/s}$ at $45^{\circ}$ from the vertical means it is also at $45^{\circ}$ from the horizontal. Thus,$u_x = u

Vedclass · Accepted Answer

$0.5, 7.5$

Vedclass · Answer

(A) Initial velocity $u = 5\sqrt{2} 	ext{ m/s}$ at $45^{\circ}$ from the vertical means it is also at $45^{\circ}$ from the horizontal. Thus,$u_x = u \cos 45^{\circ} = 5\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 5 	ext{ m/s}$ and $u_y = u \sin 45^{\circ} = 5\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 5 	ext{ m/s}$.Range $R = \frac{2 u_x u_y}{g} = \frac{2 	imes 5 	imes 5}{10} = 5 	ext{ m}$.Time of flight $T = \frac{2 u_y}{g} = \frac{2 	imes 5}{10} = 1 	ext{ s}$.The projectile splits at the highest point,which occurs at time $T/2 = 0.5 	ext{ s}$ and horizontal distance $R/2 = 2.5 	ext{ m}$.One part falls vertically,meaning its horizontal velocity becomes $0$. It takes $0.5 	ext{ s}$ to reach the ground,so $t = 0.5 	ext{ s}$.By conservation of linear momentum in the horizontal direction: $M u_x = (M/2) v_1 + (M/2) v_2$. Since the first part falls vertically,$v_1 = 0$. Thus,$M(5) = (M/2) v_2 \Rightarrow v_2 = 10 	ext{ m/s}$.The second part travels horizontally with velocity $10 	ext{ m/s}$ for $t = 0.5 	ext{ s}$ from the position $R/2 = 2.5 	ext{ m}$.Horizontal distance covered by the second part after splitting $= v_2 	imes t = 10 	imes 0.5 = 5 	ext{ m}$.Total distance $x$ from $O = (R/2) + 5 = 2.5 + 5 = 7.5 	ext{ m}$.Therefore,$t = 0.5 	ext{ s}$ and $x = 7.5 	ext{ m}$.

Similar Questions

$A$ balloon of mass $m$ is descending down with an acceleration $\frac{g}{2}$. How much mass should be removed from it so that it starts moving up with the same acceleration?

What is the ratio of the accelerations of both systems?

$A$ body of mass $2 \ kg$ is acted upon by two forces each of magnitude $1 \ N$ and inclined at $60^{\circ}$ with each other. The acceleration of the body in $m/s^2$ is $[\cos 60^{\circ}=0.5]$

What is the value of the contact force $R$ on an object of mass $m$ placed on an inclined plane with angle $\theta$?

$A$ $100 \, kg$ gun fires a ball of $1 \, kg$ horizontally from a cliff of height $500 \, m$. It falls on the ground at a distance of $400 \, m$ from the bottom of the cliff. Find the recoil velocity of the gun. (Acceleration due to gravity $g = 10 \, m/s^2$) (in $, m/s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module