Question

A stone is projected from level ground at t = 0 sec such that its horizontal and vertical components of initial velocity are 10 m/s and 20 m/s respectively. Then the instant of time at which tangential and normal components of acceleration of stone are same is: (neglect air resistance) g=10m/s2.

• A. 1 sec
• B. 2 sec
• C. 3 sec
• D. 4 sec

Answer 1

Answer: (A), (C)

The tangential and normal components of acceleration are equal at $t = 1$ sec and $t=3$ sec.

Answer 2

Solution:

For a projectile, the only acceleration is gravity:

$$\vec{a} = (0, -g) = (0, -10)$$

and the velocity components are:

$$v_x = 10,\quad v_y = 20-10t.$$

The magnitude of the velocity is:

$$|\vec{v}| = \sqrt{10^2+(20-10t)^2} = \sqrt{100+(20-10t)^2}.$$

The normal (centripetal) acceleration is given by:

$$a_n=\frac{|v_x a_y - v_y a_x|}{|\vec{v}|}=\frac{100}{\sqrt{100+(20-10t)^2}}.$$

The tangential acceleration is the rate of change of speed:

$$a_t=\frac{d|\vec{v}|}{dt}=\frac{10|20-10t|}{\sqrt{100+(20-10t)^2}}.$$

Setting $a_t = a_n$:

$$\frac{10|20-10t|}{\sqrt{100+(20-10t)^2}}=\frac{100}{\sqrt{100+(20-10t)^2}}.$$

Cancelling the common denominator:

$$10|20-10t|=100 \quad\Rightarrow\quad |20-10t|=10.$$

This gives:

$$20-10t=10 \quad \text{or} \quad 20-10t=-10.$$

Solving,

1. $20-10t=10 \Rightarrow t=1\,\text{sec}$,
2. $20-10t=-10 \Rightarrow t=3\,\text{sec}$.

Core Explanation:

Decompose velocity; use formulas $a_n=100/\sqrt{100+(20-10t)^2}$ and $a_t=10|20-10t|/\sqrt{100+(20-10t)^2}$; equate and solve $|20-10t|=10$ to get $t=1$ sec and $t=3\,$sec.