Answer
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Hint:-We have been given only two values that are the two velocities in the x and y direction. Apply the formula from the equations of kinematics ($y = {y_o} + {v_o}t + \dfrac{1}{2}a{t^2}$; where y = final distance, ${y_o}$= Initial distance; t = time; a = g = (-) acceleration). that would find the time taken to cover the distance and then put it in the formula ($s = \dfrac{d}{t}$; s = speed; d = distance; t = time).
Complete step-by-step solution:-
Step 1:
Calculate the total time taken
Apply the distance formula of kinematics,
$y = {y_o} + {v_o}t + \dfrac{1}{2}a{t^2}$ ;
Here the point where it lands the y coordinate would be zero, similarly at the starting when the ball is being hit ${y_o} = 0$.
$0 = {v_o}t + \dfrac{1}{2}a{t^2}$;
Put the given values in the above equation.
$0 = 10t + \dfrac{1}{2}( - 10){t^2}$; ..(Here $a = g = -10m/{s^2}$ a = g = -10 (indicating direction))
$0 = 10t + 5{t^2}$;
$0 = (10 + 5t)t$;
Solve the above equation
t = 0, 2 seconds;
Step2: Find the distance covered.
$s = \dfrac{d}{t}$ ;
Take t to the LHS
$d = s \times t$;
Here d = x, s = ${v_o}$, t = t.
$x = {v_o}t$;
$x = 30 \times 2$;
$x = 60m$;
Final Answer: Option “4” is correct. The ball travelled 60 m horizontally before landing on the ground.
Note:- Here it seems a question of parabolic or projectile motion as the ball is travelling horizontally and vertically and at a distance but it is not so. Here we need to apply the equation of kinematics to first find the time taken and then apply the speed, distance and time formula to find out the distance covered. Here y would be zero at the time when the ball is hit and when the ball lands.
Complete step-by-step solution:-
Step 1:
Calculate the total time taken
Apply the distance formula of kinematics,
$y = {y_o} + {v_o}t + \dfrac{1}{2}a{t^2}$ ;
Here the point where it lands the y coordinate would be zero, similarly at the starting when the ball is being hit ${y_o} = 0$.
$0 = {v_o}t + \dfrac{1}{2}a{t^2}$;
Put the given values in the above equation.
$0 = 10t + \dfrac{1}{2}( - 10){t^2}$; ..(Here $a = g = -10m/{s^2}$ a = g = -10 (indicating direction))
$0 = 10t + 5{t^2}$;
$0 = (10 + 5t)t$;
Solve the above equation
t = 0, 2 seconds;
Step2: Find the distance covered.
$s = \dfrac{d}{t}$ ;
Take t to the LHS
$d = s \times t$;
Here d = x, s = ${v_o}$, t = t.
$x = {v_o}t$;
$x = 30 \times 2$;
$x = 60m$;
Final Answer: Option “4” is correct. The ball travelled 60 m horizontally before landing on the ground.
Note:- Here it seems a question of parabolic or projectile motion as the ball is travelling horizontally and vertically and at a distance but it is not so. Here we need to apply the equation of kinematics to first find the time taken and then apply the speed, distance and time formula to find out the distance covered. Here y would be zero at the time when the ball is hit and when the ball lands.
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