Question

Isaac throws an apple straight up (in the positive direction) from 1.0 m abov
the ground, reaching a maximum height of 35 meters. Neglecting air
resistance, what is the ball's velocity when it hits the ground?​

Answer 1

Answer:

Eplanation:

Consider that when heigt is 35 meters, the velocity is zero. The you can use the formula for free fall with initial velocity zero

Vf = √(2gh) = √(2*9.8m/s^2 * 35m) = √686m^2/s^2 = 26.2 m/s

Assumption: 35 m is the maximum height from the ground, else, if the maximum heigth from the ground is 35 m + 1 m = 36 m, the numbers are:

Vf = √(2*9.8m/s^2*36m) = 26.6 m/s

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Answer 2

Therefore the ball hits the ground with 26.58 m/s.

Given:

Issac throws the ball standing on the platform which is 1 m above the ground.

The maximum height reached by the ball from the platform = 35 m

So the total distance traveled by the ball = 35 + 1 = 36 m

To Find:

The velocity of the ball when it hits the ground ( v ).

Solution:

The given question can be solved very easily as shown below.

Given that,

The ball reaches the maximum height,

So initial velocity = u = 0 m/s

Total distance traveled = s = 36 m

Acceleration due to gravity = g = 9.81 m/s² ( g = +ve because ball travels downward )

By using the formula of Kinematics,

⇒ v² - u² = 2gs

On substituting the values in the equation,

⇒ v² - 0² = 2 × 9.81 × 36

⇒ v = √( 2 × 9.81 × 36 )

⇒ v = √ 706.32 = 26.58

Therefore the ball hits the ground with 26.58 m/s.