Calculus Velocity Question

Question

What is the velocity of an object dropped from a height of 150 m when it hits the ground? Use 𝑔=−9.8m/s2 for the acceleration of gravity.

Answer 1

Velocity is the time derivative of position so we can write v = dy/dt. Likewise, acceleration is the time derivative of velocity, so a = dv/dt. In this case the acceration is g, so g = dv/dt. Rearranging the terms, we get g * dt = dv, and integrating both sides (keeping in mind that g is a constant) we get that v = g * t + C for some constant C. We can solve for t by saying that t = 0 is the time when the object is dropped, so the object initially at t = 0 has velocity of 0. Therefore we get 0 = g * 0 + C, so C = 0 and v = g * t. We can now plug this into out first equation to get g * t = dy/dt, then we rearrange the terms to get g * t * dt = dy. Integrating both sides we get (g * t^2) / 2 + C = y. Using the fact that at time t = 0, the object is at y = 150, we get C = 150, so y = (g * t^2) / 2 + 150. We can now use this equation to solve for what time the object will hit the ground (y = 0). This gives us 0 = (-9.8 * t^2) / 2 + 150, so t = sqrt(300/9.8). Plugging this back into our equation for v, we get v = -9.8 * sqrt(300/9.8), so v = -sqrt(2940).

Therefore the final velocity when the ball hits the ground is v = -54.22 m/s.

Hope this helps! Let me know if I need to explain anything.