An object is shot upwards, from the ground, with an initial velocity of 120 m/s. How high will it be after 4.0s?
h = Vo*t + 0.5g*t^2.
h = 120*4 - 4.9*4^2 =
Well, if an object is shot upwards with a velocity of 120 m/s, it must be in a hurry to reach somewhere!
To calculate how high it will be after 4.0 seconds, we can use some physics formulas. One formula that comes to mind is:
height = initial velocity * time + (1/2) * acceleration * time^2
In this case, we are given the initial velocity (120 m/s) and the time (4.0 seconds). But we're missing the acceleration.
Now, the acceleration in this case would be the acceleration due to gravity, which is about 9.8 m/s^2.
So let's plug in our values:
height = 120 m/s * 4.0 s + (1/2) * 9.8 m/s^2 * (4.0 s)^2
Now let's calculate it:
height = 480 m + 1/2 * 9.8 m/s^2 * 16.0 s^2
height = 480 m + 78.4 m
height = 558.4 m
So after 4.0 seconds, our object will be approximately 558.4 meters high! That's pretty impressive!
To find out how high the object will be after 4.0 seconds, we can use the kinematic equation for free fall:
$h = h_0 + v_0t - \frac{1}{2}gt^2$
where:
h is the height of the object after time t,
h₀ is the initial height of the object (which in this case is 0 since it is shot upwards from the ground),
v₀ is the initial velocity of the object,
t is the time, and
g is the acceleration due to gravity (-9.8 m/s²).
Given:
v₀ = 120 m/s (initial velocity),
t = 4.0 s (time).
Now, let's plug in these values into the equation:
$h = 0 + (120 m/s)(4.0 s) - \frac{1}{2}(9.8 m/s^2)(4.0 s)^2$
Simplifying:
$h = 480 m - \frac{1}{2}(9.8 m/s^2)(16 s^2)$
$h = 480 m - 78.4 m$
$h = 401.6 m$
Therefore, the object will be approximately 401.6 meters high after 4.0 seconds.