Projectiles on an Inclined Plane
PHXI04:MOTION IN A PLANE

362054 Time taken by the projectile to reach from \(A\) to \(B\) is \(t\). Then the distance \(AB\) is equal to:
supporting img

1 \(\frac{{\sqrt 3 ut}}{2}\)
2 \(\frac{{ut}}{{\sqrt 3 }}\)
3 \(2ut\)
4 \(\sqrt 3 ut\)
PHXI04:MOTION IN A PLANE

362055 A plane is inclined at an angle \(\alpha=60^{\circ}\) w.r.t. horizontal. A particle is projected with a speed \(u = 6\;m{\rm{/}}s,\) from the base of the plane, making an angle \(\theta=10^{\circ}\) w.r.t. the plane. The distance from the base, at which particle hits the plane is
(Take \(g = 10\;m{\rm{/}}{s^2},\,\sin 10^\circ = 0.17,\,\cos 10^\circ = 0.98\))

1 \(1.68\,m\)
2 \(3.64\,m\)
3 \(6.72\,m\)
4 \(2.38\,m\)
PHXI04:MOTION IN A PLANE

362056 A particle is projected from the inclined plane at angle \(37^\circ \) with the inclined plane in upward direction with speed 10 \(m\)/\(s\). The angle of inclined plane with horizontal is \(53^\circ \). Then the maximum height attained by the particle from the inclined plane will be
supporting img

1 \(4\,m\)
2 \(3\,m\)
3 \({\mathop{\rm Zero}\nolimits} \)
4 \(5\,m\)
PHXI04:MOTION IN A PLANE

362054 Time taken by the projectile to reach from \(A\) to \(B\) is \(t\). Then the distance \(AB\) is equal to:
supporting img

1 \(\frac{{\sqrt 3 ut}}{2}\)
2 \(\frac{{ut}}{{\sqrt 3 }}\)
3 \(2ut\)
4 \(\sqrt 3 ut\)
PHXI04:MOTION IN A PLANE

362055 A plane is inclined at an angle \(\alpha=60^{\circ}\) w.r.t. horizontal. A particle is projected with a speed \(u = 6\;m{\rm{/}}s,\) from the base of the plane, making an angle \(\theta=10^{\circ}\) w.r.t. the plane. The distance from the base, at which particle hits the plane is
(Take \(g = 10\;m{\rm{/}}{s^2},\,\sin 10^\circ = 0.17,\,\cos 10^\circ = 0.98\))

1 \(1.68\,m\)
2 \(3.64\,m\)
3 \(6.72\,m\)
4 \(2.38\,m\)
PHXI04:MOTION IN A PLANE

362056 A particle is projected from the inclined plane at angle \(37^\circ \) with the inclined plane in upward direction with speed 10 \(m\)/\(s\). The angle of inclined plane with horizontal is \(53^\circ \). Then the maximum height attained by the particle from the inclined plane will be
supporting img

1 \(4\,m\)
2 \(3\,m\)
3 \({\mathop{\rm Zero}\nolimits} \)
4 \(5\,m\)
PHXI04:MOTION IN A PLANE

362054 Time taken by the projectile to reach from \(A\) to \(B\) is \(t\). Then the distance \(AB\) is equal to:
supporting img

1 \(\frac{{\sqrt 3 ut}}{2}\)
2 \(\frac{{ut}}{{\sqrt 3 }}\)
3 \(2ut\)
4 \(\sqrt 3 ut\)
PHXI04:MOTION IN A PLANE

362055 A plane is inclined at an angle \(\alpha=60^{\circ}\) w.r.t. horizontal. A particle is projected with a speed \(u = 6\;m{\rm{/}}s,\) from the base of the plane, making an angle \(\theta=10^{\circ}\) w.r.t. the plane. The distance from the base, at which particle hits the plane is
(Take \(g = 10\;m{\rm{/}}{s^2},\,\sin 10^\circ = 0.17,\,\cos 10^\circ = 0.98\))

1 \(1.68\,m\)
2 \(3.64\,m\)
3 \(6.72\,m\)
4 \(2.38\,m\)
PHXI04:MOTION IN A PLANE

362056 A particle is projected from the inclined plane at angle \(37^\circ \) with the inclined plane in upward direction with speed 10 \(m\)/\(s\). The angle of inclined plane with horizontal is \(53^\circ \). Then the maximum height attained by the particle from the inclined plane will be
supporting img

1 \(4\,m\)
2 \(3\,m\)
3 \({\mathop{\rm Zero}\nolimits} \)
4 \(5\,m\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here