2) A diving platform or diving tower has three platforms at 10 m, 7.5 m, and 5.0 m. Some students are going to drop or throw straight down identical balls from each platform and they want the balls to reach the water with the same velocity. The ball from the highest platform will just be dropped with zero inital velocity. (a) What velocity do you need to throw the ball from the middle platform? (b) What velocity do you need to throw the ball from the bottom platform?
since the height h = Ho+Vo*t-4.9t^2,
the top ball will hit after t seconds, where
10+0t-4.9t^2 = 0
t = 1.429 seconds
its final speed is v = 0-1.429*9.8 = -14.00 m/s
The middle ball must then be thrown downward with velocity v such that
v-9.8t = -14.00
7.5+v*t-4.9t^2 = 0
v = -7 m/s
The low ball must satisfy
v-9.8t = -14.00
5.0+v*t-4.9t^2 = 0
v = -9.9 m/s
(a) To find the velocity needed to throw the ball from the middle platform, we can use the equation for the final velocity of an object in free fall:
vf = sqrt(2 * g * h)
where vf is the final velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the ball is dropped.
For the middle platform, the height is 7.5 m. Plugging in the values, we have:
vf = sqrt(2 * 9.8 * 7.5)
= sqrt(147)
≈ 12.12 m/s
Therefore, you need to throw the ball from the middle platform with a velocity of approximately 12.12 m/s.
(b) To find the velocity needed to throw the ball from the bottom platform, we can use the same equation as before, but this time with a height of 5.0 m:
vf = sqrt(2 * 9.8 * 5.0)
= sqrt(98)
≈ 9.90 m/s
Therefore, you need to throw the ball from the bottom platform with a velocity of approximately 9.90 m/s.
To find the velocity required to throw the ball from each platform, we can use the principles of projectile motion.
(a) For the ball to reach the water with the same velocity from the middle platform, we need to determine the initial velocity of the ball when thrown. Since the ball is thrown straight down, the initial velocity will have only the vertical component.
To find the initial velocity for the middle platform, we can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height (7.5 m)
u = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the height
At the highest point, the velocity is zero, so the time taken to reach the height can be found using:
v = u + gt
Where:
v = final vertical velocity (0 m/s)
Rearranging the equation, we get:
t = -u/g
Substituting this value of time in the equation of motion, we get:
7.5 = -u(-u/g) + (1/2)g(-u/g)^2
Simplifying the equation, we find:
7.5 = u^2/2g
Multiplying both sides by 2g, we get:
15g = u^2
Taking the square root of both sides, we find:
u = √(15g)
Therefore, the vertical velocity required to throw the ball from the middle platform is √(15g).
(b) Similarly, we can find the initial velocity required for the ball to reach the water with the same velocity from the bottom platform. Again, the ball is thrown straight down, so the initial velocity will have only the vertical component.
We can use the same equation of motion:
h = ut + (1/2)gt^2
Substituting the values, we get:
5 = -u(-u/g) + (1/2)g(-u/g)^2
Simplifying the equation, we find:
5 = u^2/2g
Multiplying both sides by 2g, we get:
10g = u^2
Taking the square root of both sides, we find:
u = √(10g)
Therefore, the vertical velocity required to throw the ball from the bottom platform is √(10g).