Question: A ball of mass 0.5kg is thrown vertically upward with a speed of 15m/s. What is its speed and direct...

Question

A ball of mass 0.5kg is thrown vertically upward with a speed of 15m/s. What is its speed and direction two seconds later?
A) 10m/s upwards
B) 5m/s upwards
C) Zero
D) 5m/s downward
E) 10 m/s downward

Answer 1

Solution

Here we have to apply the equations of kinematics that relates the initial velocity, distance, acceleration and time. We will get the maximum distance it covers in 2 seconds, and then we can put the value of distance in another equation of kinematics and find out its final velocity. Use $g = 10m{s^{ - 2}}$

Formula used:
$s = ut + \dfrac{1}{2}g{t^2}$ ;
Where:
S = Distance;
u = Initial Velocity;
g = Gravitational Acceleration;
t = Time

${v^2} = {u^2} + 2as$ ;
v = Final velocity;
u = Initial velocity;
a = g = Gravitational Acceleration;
s = Distance;

Complete step-by-step answer:
Step 1: Find the distance covered:
Apply the equation of kinematics and find out s.
$s = ut + \dfrac{1}{2}g{t^2}$ ;
Put the given values,
$s = 15 \times 2 + \dfrac{1}{2} \times - 10 \times {2^2}$ ;
Solve the above equation,
$s = 30 - (10 \times 2)$ ;
Do the needed calculation,
$s = 30 - 20$ ;
s = 10m;

Step 2: Find out the final velocity v.
Use another equation of kinematics that has final velocity v and distance s involved,
${v^2} = {u^2} + 2as$ ;
Put the given value in the above equation:
${v^2} = {15^2} + 2 \times - 10 \times 10$ ;
Calculate:
${v^2} = 225 - 200$ ;
Convert the square on the LHS to square root on the RHS:
$v = \sqrt {225 - 200}$ ;
Simplify the equation
$v = \sqrt {25}$ ;
$v = 5m/s$ ;

Final Answer: Option “B” is correct. The speed and direction of the ball two seconds later is $v = 5m/s$ and its direction is upwards.

Note: First find out the distance that is covered in 2seconds from the given velocity of 15m/s, to do that we need to choose two equations of kinematics that relate all the given variables together. Put the value of distance covered in the equation ${v^2} = {u^2} + 2as$ and find out the velocity. If the value of velocity is negative that means the direction is downwards if it is positive then it is upwards. Use: $g = - 10m{s^{ - 2}}$ .