A ball P 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

(A) The potential energy at the highest point is given by $PE = mgh$, where $h$ is the maximum height reached.For ball $P$, the angle with the vertica

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(A) The potential energy at the highest point is given by $PE = mgh$, where $h$ is the maximum height reached.For ball $P$, the angle with the vertical is $60^{\circ}$, so the angle with the horizontal is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.The maximum height for ball $P$ is $H_P = \frac{u^2 \sin^2(30^{\circ})}{2g} = \frac{u^2 (0.5)^2}{2g} = \frac{u^2}{8g}$.For ball $Q$, which is projected vertically upwards, the angle with the horizontal is $90^{\circ}$.The maximum height for ball $Q$ is $H_Q = \frac{u^2 \sin^2(90^{\circ})}{2g} = \frac{u^2}{2g}$.The ratio of potential energy of ball $P$ to ball $Q$ is $\frac{PE_P}{PE_Q} = \frac{mgH_P}{mgH_Q} = \frac{H_P}{H_Q} = \frac{u^2/8g}{u^2/2g} = \frac{2}{8} = \frac{1}{4}$.

Similar Questions

$A$ bullet is fired from a gun at the speed of $280\,m/s$ in the direction $30^{\circ}$ above the horizontal. The maximum height attained by the bullet is $........\,m$ $\left(g=9.8\,m/s^2, \sin 30^{\circ}=0.5\right):-$

$A$ cricketer can throw a ball to a maximum horizontal distance of $100\, m$. The speed with which he throws the ball is ......... $m/s$ (to the nearest integer).

$A$ bullet is fired from a cannon with a velocity of $500 \, m/s$. If the angle of projection is $15^\circ$ and $g = 10 \, m/s^2$,then the horizontal range is:

The maximum horizontal range of a projectile is $400\, m$. The maximum height attained by it will be ......... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module