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?.

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

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=

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

2.v=0

Using the formula and substitute the values

Where g is negative because motion of ball is against gravity

3.

Using the formula

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

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

5.According to question

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

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)