Question

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is upwards. Part APart complete What is the velocity of the ball when it reaches its highest point? Express your answer to two significant figures and include the appropriate units. v = 0 ms Previous Answers Correct Part B What is its velocity 0.70 s before it reaches its highest point?

Answer 1

The velocity at the highest point is 0 m/s. To find the velocity 0.70s before the highest point, we use the equation v = u + at, with a being -9.8 m/s² (gravity), and the time being -0.70s to calculate the velocity as 6.86 m/s upwards.

Step-by-step explanation:

The velocity of the ball when it reaches its highest point is 0 m/s, as all upward motion has ceased before the ball begins to fall back down due to gravity. To determine the velocity of the ball 0.70 s before it reaches its highest point, we can use the kinematic equation v = u + at, where u is the initial velocity, v is the final velocity, a is acceleration, and t is time.

Since we are given that the positive direction is upwards and acceleration due to gravity is downwards, we'll take gravity's acceleration as -9.8 m/s² (negative because it's in the opposite direction of our positive direction).

The final velocity v at the highest point is 0 m/s. We are looking for the velocity 0.70 seconds before that point, so t would be -0.70 s (negative because we are looking back in time), and we can then calculate the velocity using the equation:

v = 0 m/s + (-9.8 m/s²) × (-0.70 s)

v = 6.86 m/s

Thus, the velocity of the ball 0.70 s before it reaches its highest point is 6.86 m/s upward.

Answer 2

a) v= 0 m/s b) v= 6.86 m/s

Step-by-step explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s