the time of flight of an object projected with speed 20m/s at an angle 30¡ã with the horizontal is

Just do the vertical part of the problem.

Vi = 20 sin30 = 10/m/s up

v = Vi - g t = 10 - 9.81 t
so t is about one second upward until v = 0 at the top.
then another second falling for a total of two seconds in the air.

To find the time of flight of an object projected at an angle with the horizontal, we can use the formula:

\(t = \frac{2u \sin(\theta)}{g}\),

where:
t = time of flight
u = initial velocity of the object
θ = angle of projection
g = acceleration due to gravity

Given:
u = 20 m/s
θ = 30°
g = 9.8 m/s² (approximate value)

Substituting the values into the formula, we have:

\(t = \frac{2 \times 20 \times \sin(30°)}{9.8}\),

First, let's calculate the value of sin(30°):
sin(30°) = 0.5

Now, substituting the values into the formula:

\(t = \frac{2 \times 20 \times 0.5}{9.8}\),

Simplifying further:

\(t = \frac{20}{9.8}\).

Calculating:

\(t \approx 2.04\) seconds (rounded to two decimal places).

Therefore, the time of flight of the object is approximately 2.04 seconds.

To find the time of flight of an object projected at an angle with the horizontal, we need to consider the motion in the horizontal and vertical directions separately.

In the horizontal direction:
The initial velocity in the horizontal direction is given by:
Vx = V * cos(theta)
where V is the initial speed of the object and theta is the launch angle.

In this case, V = 20 m/s and theta = 30 degrees, so:
Vx = 20 * cos(30)
Vx = 20 * √3/2
Vx = 10√3 m/s

The horizontal component of velocity remains constant throughout the motion. Therefore, the time taken for the object to reach the maximum height (when its vertical velocity becomes zero) is given by:
t = (Vx) / (g)
where g is the acceleration due to gravity (9.8 m/s²).

In this case:
t = 10√3 / 9.8 ≈ 1.03 seconds

In the vertical direction:
The initial velocity in the vertical direction is given by:
Vy = V * sin(theta)
where V is the initial speed of the object and theta is the launch angle.

In this case, V = 20 m/s and theta = 30 degrees, so:
Vy = 20 * sin(30)
Vy = 20 * 1/2
Vy = 10 m/s

The time of flight for the object is twice the time it takes to reach maximum height, as it takes the same amount of time to fall back to the ground. Therefore, the time of flight (T) can be calculated as:
T = 2t
T = 2 * 1.03
T ≈ 2.06 seconds

So, the time of flight of the object projected at an angle of 30 degrees with the horizontal and a speed of 20 m/s is approximately 2.06 seconds.