A man stands at the top of a 27-m high building and throws a penny straight downward. If the penny has an initial velocity of -3.9 m/s, what is the velocity of the penny in m/s when it hits the ground?
A man stands at the top of a 27-m high building and throws a penny straight downward. If the penny has an initial velocity of -3.9 m/s, what is the velocity of the penny in m/s when it hits the ground?
Assumption: no air resistance.
Since the downwards direction is stated as negative, acceleration due to gravity is -10m/s2. using a=(v-u)/t, a being -10, u being initial velocity, v as velocity when hits the ground, t as time, u get t=(3.9-v)/10. find another value of such with 27=0.5 x (v+u) x t. get this equation and sub it to the previous one which is t=... to get the ans.
To find the velocity of the penny when it hits the ground, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf is the final velocity
vi is the initial velocity
a is the acceleration (which is equal to -9.8 m/s^2, as gravity is pulling the penny downward)
d is the distance travelled (which is equal to the height of the building, 27 m in this case)
Given that the initial velocity (vi) is -3.9 m/s, the acceleration (a) is -9.8 m/s^2, and the distance (d) is 27 m, we can now calculate the final velocity (vf).
Substituting the values into the equation:
vf^2 = (-3.9 m/s)^2 + 2 * (-9.8 m/s^2) * (27 m)
Simplifying the equation:
vf^2 = 15.21 m^2/s^2 + (-529.2 m^2/s^2)
vf^2 = -513.99 m^2/s^2
To find the final velocity (vf), we need to take the square root of both sides:
vf = √(-513.99 m^2/s^2)
However, we encounter a problem because the square root of a negative number is not defined in the real number system. This means that the penny will never hit the ground due to the negative initial velocity (-3.9 m/s) and the downward acceleration (-9.8 m/s^2).
Therefore, the velocity of the penny when it hits the ground is undefined in this scenario.