A projectile was launched from the ground with a certain initial velocity. Militaries used a radar to determine the vertical coordinate y(t) of the projectile for two moments of time t measured in seconds from the moment when the projectile was launched. The radar measurements showed that y(2)=269 meters,y(5)=565 meters. Calculate the maximum of y(t) if it is known as follows:

Question

A projectile was launched from the ground with a certain initial velocity. Militaries used a radar to determine the vertical coordinate y(t) of the projectile for two moments of time t measured in seconds from the moment when the projectile was launched. The radar measurements showed that y(2)=269 meters,y(5)=565 meters. Calculate the maximum of y(t) if it is known as follows:

The projectile was moving along a vertical line.
The acceleration due to gravity g is 9.81 meter/〖second〗^2.
There is a resistance proportional to the velocity of the projectile.
The value of the empirical coefficient ρ is a constant.
Distances are measured in meters.

Answer 1

try

a dy/dt + d^2 y/dt^2 = - 9.81
try
y = -(9.81/a) t + [ (V + 9.81/a)/a ] e^-at

Answer 2

To calculate the maximum of y(t), you need to determine the equation of motion for the projectile. The equation of motion is given by:

y(t) = ut - (1/2)gt² + v0t + c

Where:
- y(t) represents the vertical coordinate of the projectile at time t.
- u is the initial velocity of the projectile.
- g is the acceleration due to gravity.
- t is the time in seconds.
- v0 is the resistance proportional to the velocity of the projectile.
- c is a constant.

Since the projectile was launched from the ground, the initial vertical coordinate (y(0)) would be 0. Therefore, the equation of motion can be rewritten as:

y(t) = ut - (1/2)gt² + v0t

To find the maximum of y(t), we can differentiate the equation with respect to time and equate it to zero:

dy(t)/dt = u - gt + v0 = 0

Solving this equation for t will give us the time at which the projectile reaches its maximum height. Once we find this value of t, we can substitute it back into the equation of motion to calculate the maximum vertical coordinate (y).

Let's find u and v0 using the given information:
- y(2) = 269 meters
- y(5) = 565 meters

Using the equation of motion, we can set up two equations:
1. 269 = 2u - (1/2)g(2²) + v0(2)
2. 565 = 5u - (1/2)g(5²) + v0(5)

Now we have two equations with two variables (u and v0). We can solve these equations simultaneously to find the values of u and v0.

Once we have u and v0, we can substitute them back into the equation of motion to find the maximum vertical coordinate y(t) at the corresponding time t.

Note: The given information does not provide any details about the empirical coefficient ρ, so we do not need it to solve this problem.