A ball is thrown at an angle theta and anothe | vedclass

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

Question

(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first ball,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the

Vedclass · Accepted Answer

$15\,m, 65\,m$

Vedclass · Answer

(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first ball,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the second ball,$H_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Given $u = 40\,m/s$ and $g = 10\,m/s^2$,we have $H_1 = \frac{1600 \sin^2 	heta}{20} = 80 \sin^2 	heta$ and $H_2 = 80 \cos^2 	heta$.Given $H_2 - H_1 = 50\,m$,so $80(\cos^2 	heta - \sin^2 	heta) = 50$.Using $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$,we get $80 \cos(2	heta) = 50$,so $\cos(2	heta) = \frac{5}{8} = 0.625$.Since $\cos(2	heta) = 1 - 2\sin^2 	heta$,then $2\sin^2 	heta = 1 - 0.625 = 0.375$,so $\sin^2 	heta = 0.1875$.Thus,$H_1 = 80 	imes 0.1875 = 15\,m$.Then $H_2 = H_1 + 50 = 15 + 50 = 65\,m$.

$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.

Similar Questions

$A$ body of mass $1 \, kg$ is projected from the ground at an angle of $30^{\circ}$ with the horizontal on level ground at a speed of $50 \, m/s$. The magnitude of the change in momentum of the body during its entire flight is ....... $kg \cdot m/s$ $(g = 10 \, m/s^2)$.

The range of a projectile when launched at angle $\theta$ is same as when launched at angle $2\theta$. What is the value of $\theta$ (in $^{\circ}$)?

Two projectiles,one fired from the surface of the Earth with velocity $10 \, m/s$ and the other fired from the surface of another planet with initial speed $5 \, m/s$,trace identical trajectories. The value of the acceleration due to gravity on the planet is ......... $m/s^2$.

The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module