# Top 3 find the velocity of the coin when it gets to 50 ft in 2022 – Gấu Đây

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Date Submitted: 10/12/2019 04:25 AM

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Summary: The final velocity of the rock is -44.3m/s, and its fall time is 4.52s. There are at least two ways to solve this, one being with kinematics and the other with a combination of kinematics and energy conservation. These are just the first two that come to mind. I’ll give an explanation of both methods. Method 1: Kinematics This is a projectile motion problem which can be solved using kinematics. Because the rock is dropped from rest, we know its initial velocity is 0 (v_i=0). We are also given that it is dropped from a height of 100m (y_i=100m). Because the object hits the ground, we can define its final height as 0 (y_f=0). Finally, because the object experiences free-fall, we know that its acceleration is equal to -g, or -9.8m/s^2. We can use this kinematic equation to solve for the final velocity, v_f: v_f^2=v_i^2+2a_yΔy where Δy=y_f-y_i. Using v_i=0 as determined above, solving for v_f gives: v_f=sqrt(2a_yΔy) Using our known values: v_f=sqrt(2(-9.8m/s^2)(0m-100m)) v_f=44.3m/s downward or -44.4m/s. To find the fall time, we can use this kinematic equation: y_f=y_i+v_(iy)Δt+1/2a_yΔt^2 Because y_f and v_(iy) are both 0, we can rearrange to solve for Δt: Δt=sqrt((-2y_i)/a_y) Δt=sqrt((-2(100m))/(-9.8m/s^2) Δt=4.52s You could also use v_f=v_i+a_yΔt after finding the final velocity. Method 2: Energy Conservation/Kinematics U_(gi)+K_i=U_(gf)+K_f Where U_g is the initial gravitational potential energy (initial and final), and K is the kinetic energy (initial and final). Kinetic energy is given by K=1/2mv^2 and gravitational potential energy is given by U_g=mgh. As the rock is not moving initially (i.e. at rest), for our intents and purposes it possesses only gravitational potential energy. When it is dropped, that gravitational potential energy is transformed into kinetic potential energy as it falls. Just before the rock hits the ground, it has only kinetic energy (h≈0). Therefore, our equation becomes: U_(gi)=K_f mgh_i=1/2mv_f^2 We can see that mass cancels, as it is present on both sides, giving: gh_i=1/2v_f^2 Solving for v_f, v_f=sqrt(2gh_i) v_f=sqrt(2(9.8m/s^2)(100m)) v_f=44.3m/s You would then take that final velocity and use a kinematic equation to find Δt, either the one used above in the kinematics method or v_f=v_i+a_yΔt. Hope that helps!

Match with the search results: It’s range is + or – 8 kph (~5 mph). It will achieve that speed in about 15 meters (~50 ft). It will achieve that speed in about 3 seconds. The formula is ……. read more 