A ball of mass M is thrown vertically upwards. Another ball of mass 2 M is thrown at an angle θ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

Question

A ball of mass M is thrown vertically upwards. Another ball of mass 2 M is thrown at an angle θ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio (A) 1: 2 (B) 2: 1 (C) 1: 1 (D) 1: cos θ

Tardigrade Admin · Accepted Answer

For the ball thrown vertically upwards, the time taken by the ball to come back is

T_{1} = \frac{2 u _{1}}{g}

.
For the ball projected at an angle

θ

with the vertical, the time of flight is

T_{2} = \frac{2 u _{2} c o s θ}{g}

Since time for both the balls is same, so

\frac{2 u _{1}}{g} = \frac{2 u _{2} c o s θ}{g}

or

u_{1} = u_{2} cos θ

Now,

h_{1} = \frac{u _{1}^{2}}{2 g}

and

h_{2} = \frac{u _{2}^{2}}{2 g} cos^{2} θ

Hence,

\frac{h _{1}}{h _{2}} = \frac{u _{1}^{2}}{u _{2}^{2} c o s ^{2} θ}

= \frac{u _{2}^{2} c o s ^{2} θ}{u _{2}^{2} c o s ^{2} θ} = 1

(Using (i))

Q. A ball of mass $M$ is thrown vertically upwards. Another ball of mass $2 M$ is thrown at an angle $θ$ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

1: 2

2: 1

1: 1

$1 : cos θ$

Q. A ball of mass M is thrown vertically upwards. Another ball of mass 2M is thrown at an angle θ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

1: 2

2: 1

1: 1

1:cosθ

Q. A ball of mass $M$ is thrown vertically upwards. Another ball of mass $2 M$ is thrown at an angle $θ$ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

$1 : cos θ$