A soccer player kicks a rock with an initial velocity of 0.50 m/s from a height of 4.0 meters. What is the speed of the ball 0.70 second after it is released?
Vi = 0.50 m;s
h = 4 m
t = 0.70 s
You know the intial horizontal velocity.
the final vertical velocity is..
vfv=gt find that, then add it to the horizontal velocity as a VECTOR.
vf=sqrt(vfv^2 + Vihorz^2)
A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the horizontal velocity component of the ball to the nearest tenth of a m/s?
A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Round the answer to the nearest tenth of a second.
To find the speed of the ball 0.70 seconds after it is released, we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:
Vf = Vi + at
Where:
Vf = final velocity (speed of the ball)
Vi = initial velocity (0.50 m/s)
a = acceleration (which is equal to the acceleration due to gravity, -9.8 m/s^2)
t = time (0.70 s)
Initially, we need to determine the displacement (change in height) of the ball in order to calculate the time it takes to reach the final velocity. Since the ball is being kicked vertically upwards, we can use the equation of motion for displacement:
h = Vit + (1/2)at^2
Where:
h = displacement (4 meters)
Vi = initial velocity (0.50 m/s)
a = acceleration (which is equal to the acceleration due to gravity, -9.8 m/s^2)
t = time
Rearranging the equation, we have:
4 = 0.50t - (4.9t^2)
4.9t^2 - 0.50t + 4 = 0
We can solve this quadratic equation for t using the quadratic formula. But the roots of this equation will give us two possible values for t - when the ball is initially released and when it reaches the same height after coming down from its peak. Since we are interested in the time when the ball is at 0.70 seconds after it is released, we will discard the other root.
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
a = 4.9
b = -0.50
c = 4
t = (-(-0.50) ± sqrt((-0.50)^2 - 4 * 4.9 * 4)) / (2 * 4.9)
t = (0.50 ± sqrt(0.25 - 78.4)) / 9.8
Since 0.25 is much smaller than 78.4, we can neglect it when taking the square root. We will use the positive root to get a positive time:
t ≈ (0.50 ± sqrt(-78)) / 9.8
The square root of a negative number is not a real number, so this means that there is no positive real root for t in this equation. Therefore, the ball does not reach the same height again within 0.70 seconds after it is released.
Since the ball does not reach the same height again, this means that the time of 0.70 seconds is greater than the total time it takes for the ball to reach its peak and come back down. Therefore, we can use the initial velocity and time to calculate the final velocity using the first kinematic equation:
Vf = Vi + at
Vf = 0.50 m/s + (-9.8 m/s^2)(0.70 s)
Vf = 0.50 m/s - 6.86 m/s
Vf ≈ -6.36 m/s
The negative sign indicates that the ball is moving downwards. Therefore, the speed of the ball 0.70 seconds after it is released is approximately 6.36 m/s downwards.