Neglecting air resistance, how fast is the object traveling when it strikes the gounnd at the end of 3.8s?

v=gt=9.8•3.8=…

To calculate the speed of an object neglecting air resistance, we can use the equations of motion. Since we only know the time and neglect air resistance, we need to consider the motion in the vertical direction, assuming the object is only falling.

We can use the equation: s = ut + (1/2)at^2, where,
s = distance traveled by the object (height)
u = initial velocity of the object (0 m/s, as it starts from rest)
t = time taken (3.8 s)
a = acceleration (acceleration due to gravity, g = 9.8 m/s^2)

In this case, the equation becomes: s = (1/2)gt^2

To find the speed at the end of 3.8 seconds, we need to determine the distance traveled by the object. Assuming the object is falling vertically downward, the distance traveled is the height it has fallen.

Using the equation: s = (1/2)gt^2, we can calculate the distance traveled:
s = (1/2)(9.8 m/s^2)(3.8 s)^2
s = (1/2)(9.8 m/s^2)(14.44 s^2)
s ≈ 70.828 m

Now, since the object falls freely, it reaches its maximum speed when it hits the ground. Therefore, the speed at the end of 3.8 seconds is equal to the final velocity when it hits the ground.

To calculate the final velocity, we can use the equation: v^2 = u^2 + 2as, where
v = final velocity
u = initial velocity (0 m/s)
a = acceleration (acceleration due to gravity, g)
s = distance traveled by the object (height)

Rearranging the equation, we have: v = sqrt(u^2 + 2as)

Substituting the given values:
v = sqrt((0 m/s)^2 + 2(9.8 m/s^2)(70.828 m))
v = sqrt(2(9.8 m/s^2)(70.828 m))
v = sqrt(1377.216 m^2/s^2)
v ≈ 37.12 m/s

Therefore, neglecting air resistance, the object will be traveling at approximately 37.12 m/s when it strikes the ground at the end of 3.8 seconds.