A SUPERSONIC FIGHTER JET FLYING HORIZONTALLW AT HEIGHT OF 4KM ABOVE THE GROUND AND AT A SPEED OF 400M/S FIRES A ROCKET VERTICALLY DOWNWARD. THE ROCKET IS RELEASED WITH AN INITIAL SPEED OF 80M/S CALCULATE THE TIME OF FLIGHT OF THE ROCKET AND THE HORIZONTAL DISTANCE FROM THE POINT OF ITS RELEASE AT WHICH THE ROCKET LANDS
No need to shout.
4000 = 80t + 4.9 t^2
You'll have to solve the quadratic
Horizontal will be 400t
To calculate the time of flight of the rocket and the horizontal distance it travels, we can use the equations of motion.
Let's break down the problem into the horizontal and vertical components.
Horizontal Motion:
The horizontal component of motion remains at a constant speed of 400 m/s, so the time of flight (t) can be calculated by dividing the horizontal distance (d) by the horizontal velocity (v):
t = d / v
Vertical Motion:
The rocket is initially at a height of 4 km above the ground (h). The initial vertical velocity (u) is 80 m/s, and the final vertical velocity (v) will be -80 m/s due to gravity acting downward. We need to find the time it takes for the rocket to reach the ground.
Using the equation of motion for vertical motion:
h = ut + (1/2)gt^2
where:
u = initial vertical velocity
g = acceleration due to gravity (assumed to be 9.8 m/s^2)
We can rearrange the equation to find the time of flight:
t = sqrt((2h) / g)
Now let's substitute the values:
d = ?
v = 400 m/s
h = 4 km = 4000 m
u = 80 m/s
g = 9.8 m/s^2
Calculating the time of flight:
t = sqrt((2 * 4000) / 9.8)
t ≈ 20.20 seconds
To find the horizontal distance traveled (d), we can use the equation:
d = v * t
d = 400 * 20.20
d ≈ 8080 meters
Therefore, the time of flight of the rocket is approximately 20.20 seconds, and it lands at a horizontal distance of approximately 8080 meters from the point of its release.