How High Does a 64 lb Projectile Go When Fired Upward at 2000 ft/sec?

[anik18]

A projectile weighing 64 lb is fired directly upward with a velocity 2000 ft/sec. Assume that the accerleration due to gravity is constant, and the air resists the motion with a force proportional to the velocity. When the velocity is 1000 ft/sec, the resisting force is 20 lb. Find the maximum height attained by the projectile, and the time required for it to reach this height.

please help this.
thankyou

 

[D H]

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[tacman]

To solve this problem, we can use the equation of motion for a projectile:

m(d^2y/dt^2) = -mg - bv

Where m is the mass of the projectile, g is the acceleration due to gravity, b is the proportionality constant for air resistance, and v is the velocity.

We know that the mass of the projectile is 64 lb and the acceleration due to gravity is -32 ft/sec^2 (since it is acting in the opposite direction of the projectile's motion). We also know that when the velocity is 1000 ft/sec, the resisting force is 20 lb. Therefore, we can set up the following equation:

64(-32) = -64b(1000) - 20

Solving for b, we get b = 0.0005 lb/sec. Now, we can use this value of b in the equation of motion to find the maximum height attained by the projectile:

m(d^2y/dt^2) = -mg - bv

64(d^2y/dt^2) = -64(-32) - 0.0005(2000)

d^2y/dt^2 = 32 + 0.0005(2000)

d^2y/dt^2 = 32 + 1

d^2y/dt^2 = 33

Integrating twice with respect to time, we get:

y = 16t^2 + C1t + C2

Where C1 and C2 are constants of integration. We can use the initial conditions given in the problem to find the values of C1 and C2:

When t = 0, y = 0 (since the projectile is fired from the ground)

Therefore, C2 = 0.

When v = 0, y = h (maximum height attained by the projectile)

Therefore, 0 = 16(t^2) + C1t

Using the given velocity of 2000 ft/sec at t = 0, we can solve for C1:

2000 = 16(0^2) + C1(0)

C1 = 2000

Therefore, our equation becomes:

y = 16t^2 + 2000t

To find the time required for the projectile to reach its maximum height, we can use the fact that at the maximum height, the velocity is 0