A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s²)

Concept:

• It is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
• Initial Velocity: The initial velocity can be given as x components and y components.

• Component of initial velocity in x-direction, (ux) = u cos θ
• Component of initial velocity in the y-direction, (uy) = u sin θ
• In the case of projectile motion, we can see a free-fall motion of a body on a parabolic path with constant velocity.
• If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below.

• And thus, the range of a projectile is the displacement of a particle along the x-axis and can be given as:

The height of projectile is given by,

\(H_2~=~\frac{u^2_y}{2g}\)

• Whereas the time of flight is the total time for which projectile stayed in the air.

Time of flight for the projectile,

\(\left( {\bf{t}} \right) = \;\frac{{2 ~u ~sin~θ }}{g}~=~\frac{2u_y}{g}\)

where The angle of projection = θ, Initial velocity = u, Gravitational acceleration = g, Time of flight = t, Range of projectile = R

Calculation:

Given:

Height of the tower H₁ = 60 m, initial velocity of the object u = 40 m/s, angle of inclination with horizontal θ = 30°, u_x = 40 × cos 30° = 20√3 m/s, u_y = 40 × sin 30° = 20 m/s

Time taken to reach the maximum height is given by,

\(T_1~=~\frac{u_y}{g}~=~\frac{20}{10}~=~2~s\)

Height above the tower is given by,

\(H_2~=~\frac{u^2_y}{2g}~=~\frac{20~\times ~20}{2~\times~ 10}~=~20~m\)

Therefore, the maximum height is,

H_max = H₁ + H₂ = 60 + 20 = 80 m

Now, the time taken to free fall from maximum height is,

\(T_2~=~\sqrt{\frac{2H_{max}}{g}}~=~\sqrt{\frac{2~\times ~80}{10}}~=~4~s\)

Thus, the total time is taken during the entire flight is given by,

T_total = T₁ + T₂ = 2 + 4 = 6 s