Question

A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.2 m/s from a height of 32 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. Take up as the positive direction.

1. What is the speed of the blue ball when it reaches its maximum height?.
2. How long does it take the blue ball to reach its maximum height?.
3. What is the maximum height the blue ball reaches?.
4. What is the height of the red ball 3.77 seconds after the blue ball is thrown?.
5. How long after the blue ball is thrown are the two balls in the air at the same height?.

Answer 1

1. The speed of the blue ball when it reaches its maximum height is 24.1 m/s. 2. It takes 2.464 seconds for the blue ball to reach its maximum height. 3. The maximum height the blue ball reaches is 29.88 meters. 4. The height of the red ball 3.77 seconds after the blue ball is thrown is -12.965 meters. 5. The two balls are at the same height 1.926 seconds after the blue ball is thrown.

Step-by-step explanation:

1. The speed of the blue ball when it reaches its maximum height can be found by using the 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 of the blue ball is 24.1 m/s and the acceleration is -9.81 m/s² (taking upward as positive). Since the ball reaches its maximum height, its final velocity at that point is 0 m/s. Plugging these values into the equation, we can solve for t.

2. To find how long it takes for the blue ball to reach its maximum height, we can use the 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 of the blue ball is 24.1 m/s and the acceleration is -9.81 m/s² (taking upward as positive). Since the ball reaches its maximum height, its final velocity at that point is 0 m/s. Plugging these values into the equation, we can solve for t.

3. The maximum height the blue ball reaches can be found using the equation s = ut + 1/2at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity of the blue ball is 24.1 m/s and the acceleration is -9.81 m/s² (taking upward as positive). We want to find the displacement when the velocity of the ball is 0 m/s, so we can set v = 0 and solve for t. Once we have the value of t, we can substitute it back into the equation to find the maximum height.

4. To find the height of the red ball 3.77 seconds after the blue ball is thrown, we can use the equation s = ut + 1/2at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity of the red ball is 7.2 m/s and the acceleration is -9.81 m/s² (taking upward as positive). We want to find the displacement after 3.77 seconds, so we can plug these values into the equation to find the height.

5. The two balls will be at the same height when their displacements are equal. To find the time when this happens, we can set the equations for their displacements equal to each other and solve for t. The displacement equation for the blue ball is s = ut + 1/2at², while the displacement equation for the red ball is s = gt², where g is the acceleration due to gravity. Plugging in the given values and solving for t will give us the time when the two balls are at the same height.

Answer 2

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Step-by-step explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g=
`9.81ms^(-2)`

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

`v=u+at`

Using the formula and substitute the values

`0=24.1-9.81t`

Where g is negative because motion of ball is against gravity

`24.1=9.81t`

`t=(24.1)/(9.81)=2.46s`

3.
`y=y_0+ut+(1)/(2)at^2`

Using the formula

`y=0.5+24.1(2.46)-(1)/(2)(9.81)(2.46)^2`

`y=30.1 m`

4.Time of flight for red ball=3.77-2.9=0.87s

`y'=32-7.2(0.87)-(1)/(2)(9.81)(0.87)^2`

`y'=22.0m`

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

`0.5+24.1(t+2.9)-(1)/(2)(9.81)(2.9+t)^2=32-7.2t-(1)/(2)(9.81)t^2`

`0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2`

`0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2`

`0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t`

`-2.86105=-2.851t`

`t=(2.86105)/(2.851)=1.004 s`

Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.