a tennis ball was held 4.5 ft above the surface of a tennis court and was thrown vertically downward with an initial velocity whose magnitude was 12 fps. determine the magnitude of the velocity of the ball just before it struck the surface of the court
This is engineering? Hmmmm.
Vf^2=Vi^2 + 2 ad
Vf^2=12f/s+2*-32f/s^2*-4.5
Vf^2=288
Vf=12 sqrt2 fps
check that
the square root of 288 comes out to 16.9 so that would be my answer right? how did you come up with 12 sqrt2.
To determine the magnitude of the velocity of the ball just before it struck the surface of the court, we can use the equation of motion for an object in free fall.
The equation we'll use is:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration (which is equal to the acceleration due to gravity, approximately 32.2 ft/s^2)
s = displacement
First, let's assign the given values to the variables:
u = 12 ft/s (initial velocity)
a = 32.2 ft/s^2 (acceleration due to gravity)
s = -4.5 ft (displacement in the downward direction)
Since the ball is thrown vertically downward, the displacement is negative.
Now, we can plug these values into the equation:
v^2 = (12 ft/s)^2 + 2 * 32.2 ft/s^2 * (-4.5 ft)
Simplifying further:
v^2 = 144 ft^2/s^2 - 289.8 ft^2/s^2
v^2 = -145.8 ft^2/s^2
To find the magnitude of the velocity, we take the square root of both sides of the equation:
v = √(-145.8) ft/s
Since we cannot take the square root of a negative number, it means that the ball will never reach the surface of the court. The magnitude of the velocity just before it hits the surface will be imaginary.