a projectile is launched at 35 degrees above the horizontal with an initial velocity of 120m/s. What is the projectile's speed 3 seconds later?
Figure the new vertical velocity at t=3. then the speed as equal to sqrt(Vv^2+ Vh^2)
The other way to do this is figure how high it travels in three seconds, then use the fact that
Kineticenergy=initialKE- change in PE.
To find the projectile's speed 3 seconds later, we need to calculate its vertical and horizontal velocities at that time.
First, let's determine the vertical velocity (Vv) after 3 seconds. We know the initial vertical velocity is given by Vv = V * sin(theta), where V is the initial velocity (120 m/s) and theta is the launch angle (35 degrees above the horizontal). So, Vv = 120 m/s * sin(35 degrees).
Next, we can calculate the horizontal velocity (Vh) after 3 seconds. The horizontal velocity remains constant throughout the projectile's motion. Hence, Vh = V * cos(theta), where V is the initial velocity (120 m/s) and theta is the launch angle (35 degrees above the horizontal). Thus, Vh = 120 m/s * cos(35 degrees).
Now that we have the vertical (Vv) and horizontal (Vh) velocities, we can find the projectile's speed 3 seconds later using the formula:
Speed = sqrt(Vv^2 + Vh^2)
Substituting the values we calculated earlier, we get:
Speed = sqrt((120 m/s * sin(35 degrees))^2 + (120 m/s * cos(35 degrees))^2)
Evaluating this expression will give us the projectile's speed 3 seconds later.