At time t=0 a projectile passes through the point (0,0,192) with a velocity v=10i-2j+16k. Because of spin and gravity, the projectile experiences an acceleration given by a=2i-32k.
a. Find the velocity v(t) and position vector R(t) at time t>0.
b. At what time t does the projectile hit the ground?
c. Find the location of the projectile when it hits the ground
To find the velocity vector v(t) and position vector R(t) at time t > 0, we can use the equations of motion that relate velocity, acceleration, and time.
a. Velocity v(t):
The equation for velocity is given by:
v(t) = v(0) + ∫a(t) dt
where v(0) is the initial velocity at time t=0.
Given:
v(0) = 10i - 2j + 16k
a(t) = 2i - 32k
Integrating the acceleration over time, we get:
∫a(t) dt = ∫(2i - 32k) dt = 2t i + (-32t) k
Therefore, the velocity vector v(t) at time t>0 becomes:
v(t) = v(0) + 2t i - 32t k
Position vector R(t):
The equation for position is given by:
R(t) = R(0) + ∫v(t) dt
where R(0) is the initial position at time t=0.
Given:
R(0) = 0i + 0j + 192k (initial position)
Integrating the velocity over time, we get:
∫v(t) dt = ∫(10i - 2j + 16k + 2t i - 32t k) dt
= (10t + t^2)i + (-2t)j + (16t - 16t^2 + 192)k
Therefore, the position vector R(t) at time t>0 becomes:
R(t) = R(0) + (10t + t^2)i - 2tj + (16t - 16t^2 + 192)k
b. To find the time t when the projectile hits the ground, we need to determine when the y-component of the position vector R(t) becomes zero. Since the ground is located at y=0, we set -2t = 0 and solve for t.
-2t = 0
t = 0
Therefore, the projectile hits the ground at t = 0.
c. To find the location of the projectile when it hits the ground, we substitute t = 0 into the position vector R(t).
R(t) = R(0) + (10t + t^2)i - 2tj + (16t - 16t^2 + 192)k
R(0) = 0i + 0j + 192k
Substituting t = 0:
R(t) = 0i + 0j + 192k
Hence, the location of the projectile when it hits the ground is (0, 0, 192).