A projectile is projected at an angle 300 with the horizontal with the velocity 30 ms-1
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What is the time of flight
If you shot it at 300 degrees from horizontal you shot it into the ground.
maybe you meant 30 deg.
if 30°, then just find t when the height is zero:
30*sin30° t - 4.9t^2 = 0
To find the time of flight of a projectile, we need to consider the motion of the projectile in both the vertical and horizontal directions.
First, let's determine the initial vertical velocity (Vy) and the initial horizontal velocity (Vx).
Given:
Angle of projection (θ) = 30 degrees
Initial velocity (V) = 30 m/s
To calculate Vy and Vx, we can use the following trigonometric formulas:
Vy = V * sin(θ)
Vx = V * cos(θ)
Plugging in the values, we have:
Vy = 30 * sin(30)
Vy = 15 m/s
Vx = 30 * cos(30)
Vx = 25.98 m/s (approximately)
Next, we can calculate the time of flight (T).
The time of flight is the total time it takes for the projectile to reach the same level from which it was launched. This means the vertical displacement of the projectile when it hits the ground is zero.
Using the equation for vertical displacement:
Vertical displacement (Sy) = Vy * T + (1/2) * g * T^2
Since Sy is zero (the projectile returns to the same level), we have:
0 = Vy * T + (1/2) * g * T^2
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Simplifying the equation, we get:
(1/2) * g * T^2 = -Vy * T
(1/2) * 9.8 * T^2 = -15 * T
4.9 * T^2 = -15 * T
Dividing both sides by T:
4.9 * T = -15
T = -15 / 4.9
T ≈ -3.06 (approximately)
The negative value indicates that the time of flight is in the opposite direction from the initial vertical velocity. However, since time cannot be negative, we ignore the negative sign and take the absolute value:
T ≈ 3.06 seconds (approximately)
Therefore, the time of flight of the projectile is approximately 3.06 seconds.