Question

Two particles of same mass are projected simultaneously with the same speed $ 20m{s^{ - 1}} $ from the top of a tower of height $ 20m $ . One is projected vertically upwards and other projected horizontally. The maximum height attained by the centre of mass from the ground will be $ (g = 10m{s^{ - 2}}) $ .

Answer 1

Hint: According to this question, we need to find the maximum height attained by the centre of mass from the ground. So, this question is based on projectile motion. Another thing to keep in mind is that vertical distance will be provided by the mass projected vertically upward.

Complete step by step answer:
An object that becomes airborne after it is thrown or projected is called a projectile. Example, football and javelin throw.
Some important points regarding projectile motion are:
1) Projectile motion comprises two parts – horizontal motion of no acceleration and vertical motion of constant acceleration due to gravity.
2) Projectile motion is in the form of a parabola.
3) Projectile motion is usually calculated by neglecting air resistance to simplify calculations.
We know that the vertical distance will be provided by the mass projected vertically upward,
So at maximum height,
 $ {v_y} = 0m/s $
 $ {u_y} = 20m/s $
 $ {a_y} = - 10m/{s^2} $
Applying the third equation of motion, we get,
 $ v_y^2 - u_y^2 = 2{a_y}{s_y} $ $ $
 $ 0 - {(20)^2} = 2 \times ( - 10) \times {s_y} $
On simplifying the above equation, we get,
 $ {S_y} = 20m $
So the final answer is $ {S_y} = 20m $ .

Additional Information:
Galileo was the first person who ever accurately described projectile motion. He was the one who first broke down motion into its separate horizontal and vertical components. Galileo even took this idea further with his realization that there was more than one force at work upon the projectile.

Note:
Be careful while applying the formula. Remember that an object that becomes airborne after it is thrown or projected is called a projectile. Also remember that projectile motion is in the form of a parabola. Calculation mistakes are possible, so try to avoid them and be sure of the final answer.