Projectile motion help?

Question

1. The problem statement, all variables and given/known data A missle is fired vertically up from a ground with an initial velocity of 20m/s. The acceleration at any time is given by dv/dt = -10 + t/2 m/s^2. Find the height of the missile after two seconds Find the height after four seconds the direction the missile is traveling when t = 4

Answer 1

(1)

The initial Velocity of the missile is 20 m/sec.

Accerlation of the missile dv/dt = -10 + t/2 m/sec².

Height at time h = 2 sec.

Accerlation at t = 2sec is dv/dt = -10 + (2)/2 = -10 + 1 = -9 m/sec

We know that S = ut + (1/2)at²

S = 20 * 2 + 0.5 * (-9) * 2²

S = 40 - 18

S = 22 mts

Therefore the Height of the missile afte 2 sec is 22 mts.

Answer 2

(2)

The initial Velocity of the missile is 20 m/sec.

Accerlation of the missile dv/dt = -10 + t/2 m/sec².

Height at time t = 4 sec.

Accerlation at t = 4sec is dv/dt = -10 + (4)/2 = -10 + 2 = -8 m/sec

We know that S = ut + (1/2)at²

S = 20 * 4 + 0.5 * (-8) * 4²

S = 80 - 16

S = 16 mts

Therefore the Height of the missile afte 4 sec is 16 mts.

Answer 3

(3)

The initial Velocity of the missile is 20 m/sec.

Accerlation of the missile dv/dt = -10 + t/2 m/sec².

Height at time h = 2 sec.

Accerlation at t = 2sec is dv/dt = -10 + (2)/2 = -10 + 1 = -9 m/sec

We know that S = ut + (1/2)at²

S = 20 * 2 + 0.5 * (-9) * 2²

S = 40 - 18

S = 22 mts

The Height of the missile afte 2 sec is 22 mts.

Height at time t = 4 sec.

Accerlation at t = 4sec is dv/dt = -10 + (4)/2 = -10 + 2 = -8 m/sec

We know that S = ut + (1/2)at²

S = 20 * 4 + 0.5 * (-8) * 4²

S = 80 - 16

S = 16 mts

The Height of the missile afte 4 sec is 16 mts.

Therefore the we can observe that the height of the missile at t = 2 sec is 22 m and at time t = 4 sec is 16 mts.

The Height of the missile is decreasing from t = 2 sec to t = 4 sec.

Therefore the Direction of Missile is towards the ground.