Question

A ball is thrown vertically upwards with a given velocity v such that it rises for T seconds (T> 1). What is the distance traversed by the ball during the last one second of ascent (in meters)? (Acceleration due to gravity is g m/${{s}^{2}}$ ).
(A) $v+g\dfrac{(1-2T)}{2}$
(B) $vT+\dfrac{1}{2}g[{{T}^{2}}-{{(T-1)}^{2}}]$
(C) $\dfrac{g}{2}$
(D) $\dfrac{1}{2}g[{{T}^{2}}-{{(T-1)}^{2}}]$

Answer 1

Second equation of motion is given by:
S= $ut+\dfrac{1}{2}a{{t}^{2}}$ where,
s is the distance travelled
t is the time taken
u is the initial velocity
a is the acceleration
The acceleration experienced by a body under free fall due to the gravitational force of the massive body
Using the second equation of motion we will solve this question

Complete Step by step solution:
The upward velocity given to the ball is v.
Acceleration works on it in the downward direction due to gravity g.
Distance covered by it in T seconds,${{S}_{T}}$ =$vT-\dfrac{1}{2}g{{T}^{2}}$ ----- (1)
Distance covered by it in T−1 seconds, ${{S}_{T-1}}$=$v(T-1)-\dfrac{1}{2}g{{(T-1)}^{2}}$ ----- (2)
Hence distance covered in last second, S= ${{S}_{T}}$- ${{S}_{T-1}}$------- (3)
Putting the value of equation (1) and equation (2) in equation (3) we get,
S= $\left[ vT-\dfrac{1}{2}g{{T}^{2}} \right]-\left[ v(T-1)-\dfrac{1}{2}g{{(T-1)}^{2}} \right]$
Now let us simplify the term by using the algebraic formula for the square of sum of terms.
S= $\left[ vT-\dfrac{1}{2}g{{T}^{2}} \right]-\left[ vT-v-\dfrac{1}{2}g({{T}^{2}}+1-2T) \right]$
After simplifying the above equation further, we get
S= $v+g\dfrac{(1-2T)}{2}$

Correct option is A

Note: Any object affected only by gravity (a projectile or an object in free fall) has an acceleration of -9.81 m/s2, regardless of the direction. The acceleration is negative when going up because the speed is decreasing. The acceleration is negative when going down because it is moving in the negative direction, down. Velocity – Time graph is used to derive the equations of motion. It is clear that the value of g decreases with an increase in height of an object. Hence the value of g becomes zero at infinite distance from the earth.