A student drops an iPhone from a bridge to the water 12.0m below. with what speed does the phone strike the water
To determine the speed with which the iPhone strikes the water, we can use the equation for free fall motion:
v² = u² + 2as
where:
v is the final velocity (which we want to find),
u is the initial velocity (which is zero as the iPhone is dropped from rest),
a is the acceleration due to gravity (-9.8 m/s² for free fall near the Earth's surface),
s is the displacement (12.0 m in this case, as the iPhone falls 12.0m below).
Substituting the given values into the equation:
v² = 0 + 2(-9.8)(12.0)
v² = -235.2
v = √(-235.2)
Since velocity is a vector quantity and cannot have a negative magnitude, it indicates the direction. Therefore, we take the magnitude of the velocity:
v ≈ 15.33 m/s
Therefore, the iPhone strikes the water with a speed of approximately 15.33 m/s.
To determine the speed at which the phone strikes the water, we can use the equation for velocity, which is:
v = √(2 * g * h)
Where:
v = velocity (speed)
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height from which the phone is dropped (12.0 m)
Plugging in the values:
v = √(2 * 9.8 * 12.0)
Calculating further:
v = √(235.2)
v ≈ 15.33 m/s
Therefore, the phone will strike the water with a speed of approximately 15.33 m/s.
To find the speed with which the phone strikes the water, we can use the equations of motion.
Assuming the phone is dropped vertically, we can use the following equation to find the final velocity just before it hits the water:
v^2 = u^2 + 2as
Here, v is the final velocity, u is the initial velocity (which is zero since the phone is dropped), a is the acceleration due to gravity (approximately 9.8 m/s^2), and s is the displacement (12.0m since the phone falls 12.0m).
Plugging in the values, we get:
v^2 = 0 + 2 * 9.8 * 12.0
v^2 = 235.2
Taking the square root of both sides, we find:
v ≈ 15.3 m/s
Therefore, the phone strikes the water with a speed of approximately 15.3 m/s.