A water balloon was dropped from a high window and struck its target 3.9 seconds later. If the balloon left the person's hand at -8.4 m/sec, what was its velocity on impact?
16m/s
To determine the velocity of the water balloon on impact, we can use the equation of motion:
v = u + at
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
Given:
Initial velocity, u = -8.4 m/s
Acceleration, a = -9.8 m/s^2 (assuming it is the acceleration due to gravity)
Time, t = 3.9 seconds
Let's substitute these values into the equation to calculate the final velocity:
v = -8.4 m/s + (-9.8 m/s^2) * 3.9 s
First, multiply -9.8 m/s^2 by 3.9 s:
v = -8.4 m/s - 38.22 m/s
Then, add the two values together:
v = -46.62 m/s
Therefore, the velocity of the water balloon on impact is approximately -46.62 m/s. Note that the negative sign indicates the direction, which in this case indicates that the balloon was moving downward when it struck its target.
To find the velocity of the water balloon on impact, we can use the kinematic equation:
v = u + at
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
In this case, the initial velocity is given as -8.4 m/sec (negative sign indicates it was thrown downward), and the time of impact is given as 3.9 seconds.
However, we need to determine the acceleration first. Since the only force acting on the water balloon is gravity, we can use the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative sign indicates it acts downward).
Substituting the values into the equation, we have:
v = -8.4 m/s + (-9.8 m/s^2) * 3.9 sec
Calculating this, we get:
v = -8.4 m/s - 38.22 m/s
v ≈ -46.62 m/s
Therefore, the water balloon's velocity on impact was approximately -46.62 m/s.