the height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with initial velocity of 24.5 m/s is
h=2+24.5t-4.9t^2 after t seconds. Find the velocity after 2s and 4 s?
gotta remember that
v(t) = h'(t) = 24.5 - 9.8t
now just plug in t=2 or 4
To find the velocity after 2 seconds and 4 seconds, we need to differentiate the height equation h = 2 + 24.5t - 4.9t^2 with respect to time (t).
The formula for velocity is v = dh/dt, where dh/dt represents the derivative of height (h) with respect to time (t).
Differentiating h = 2 + 24.5t - 4.9t^2, we get:
dh/dt = d(2)/dt + d(24.5t)/dt - d(4.9t^2)/dt
= 0 + 24.5 - 2(4.9t)
= 24.5 - 9.8t
Now, we can substitute the values of t into the derivative we just found to find the velocity at those times.
1. Velocity after 2 seconds (t = 2):
v(2) = 24.5 - 9.8(2)
= 24.5 - 19.6
= 4.9 m/s
2. Velocity after 4 seconds (t = 4):
v(4) = 24.5 - 9.8(4)
= 24.5 - 39.2
= -14.7 m/s
Therefore, the velocity after 2 seconds is 4.9 m/s and the velocity after 4 seconds is -14.7 m/s.
To find the velocity after 2 seconds and 4 seconds, we need to differentiate the equation for height with respect to time. The derivative of height with respect to time gives us the velocity.
The equation for height is given as:
h = 2 + 24.5t - 4.9t^2
To find the velocity, we can differentiate this equation with respect to time (t):
dh/dt = 24.5 - 9.8t
Now, we have the equation for velocity in terms of time. To find the velocity after 2 seconds, substitute t = 2 into the equation:
v(2) = 24.5 - 9.8(2)
= 24.5 - 19.6
= 4.9 m/s
Therefore, the velocity after 2 seconds is 4.9 m/s.
Similarly, to find the velocity after 4 seconds, substitute t = 4 into the equation:
v(4) = 24.5 - 9.8(4)
= 24.5 - 39.2
= -14.7 m/s
Therefore, the velocity after 4 seconds is -14.7 m/s (since it's negative, it indicates the object is moving downward).