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Question 5

A ball is thrown straight down with an initial speed of 4.30 m/s, from a height of
8.80 m. What is the velocity of the ball when it reaches a height of 5.80?

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User Commanda
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Answer:We can use the kinematic equations of motion to solve this problem. The equation we need to use is:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (which is -9.81 m/s^2), and s is the displacement.

Initially, the ball is thrown straight down with an initial speed of 4.30 m/s and a height of 8.80 m. Let's take the upward direction as positive. Using the equation of motion for displacement, we can find the time it takes for the ball to reach a height of 5.80 m:

s = ut + (1/2)at^2

-3 = 4.3t + (1/2)(-9.81)t^2

-4.905t^2 + 4.3t - 3 = 0

Using the quadratic formula, we find that the time it takes for the ball to reach a height of 5.80 m is t = 0.956 s (rounded to three significant figures).

Now, we can use the equation of motion for velocity to find the velocity of the ball at this point:

v^2 = u^2 + 2as

v^2 = (4.3 m/s)^2 + 2(-9.81 m/s^2)(-3 m)

v = -5.08 m/s

The negative sign indicates that the velocity is in the downward direction. Therefore, the velocity of the ball when it reaches a height of 5.80 m is 5.08 m/s downward.

Step-by-step explanation:

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User RKRK
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