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A golf ball is thrown downward with a velocity of -2.97 m/s, from a height of 9.77 m, What is the impact velocity of the ball on the surface of earth?

User CrazyCasta
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Final answer:

The impact velocity of the golf ball thrown downward from a height of 9.77 m with an initial velocity of -2.97 m/s is approximately -14.14 m/s.

Step-by-step explanation:

The question asks for the impact velocity of a golf ball thrown downward with an initial velocity and from a given height. To calculate this, we can use the kinematic equations of motion, specifically the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.81 m/s^2), and s is the distance traveled.

Given the initial velocity (u) of -2.97 m/s (the negative sign indicates the direction is downwards), the height (s) from which the ball is thrown as 9.77 m, and a as 9.81 m/s^2 (positive because it's directed downwards and we've taken downwards as negative for the velocity), we can rearrange the equation to solve for v:

v^2 = (-2.97 m/s)^2 + 2(9.81 m/s^2)(9.77 m)
v^2 = 8.8209 m^2/s^2 + 191.3014 m^2/s^2v^2 = 200.1223 m^2/s^2
v = sqrt(200.1223 m^2/s^2)
v ≈ -14.14 m/s

The negative sign indicates that the ball is moving downward upon impact. So, the impact velocity of the golf ball is approximately -14.14 m/s.

User Congbaoguier
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