Different Equations of Motion for Free Falling Object

Transcript

Old Equation v = u + at s = ut + 1/2 at2 v2 = u2 + 2as where u = Initial Velocity v = Final Velocity a = Acceleration t = Time taken s = Distance New Equation v = u + gt h = ut + 1/2 gt2 v2 = u2 + 2ah where u = Initial Velocity v = Final Velocity g = Acceleration due to gravity t = Time taken h = Height of object Question To estimate the height of a bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 Seconds to touch the water surface in the river. Calculate the height of the bridge from the water level (g = 9.8 m/s2) Since the stone is dropped freely Initial velocity = u = 0 m/s Time taken = t = 2s Acceleration = a = +g = +9.8 m/s2 Height of bridge will be the distance travelled by stone i.e. s (Acceleration is positive because stone falls downwards) We know u, a and t Finding s using 2nd equation of motion. s = ut + 1/2 at2 = 0 × 2 + 1/2 × 9.8 × (2)2 = 0 + 1/2 × 9.8 × 4 = 19.6 m Hence, height of the bridge in 19.6 m We know u, a and t Finding s using 2nd equation of motion. s = ut + 1/2 at2 = 0 × 2 + 1/2 × 9.8 × (2)2 = 0 + 1/2 × 9.8 × 4 = 19.6 m Hence, height of the bridge in 19.6 m (Acceleration is negative because ball is thrown upwards) Finding Initial velocity (u) We know v, a & a, Finding u using 3rd equation of motion v2 − u2 = 2as 02 − u2 = 2 × (−9.8) × 19.6 –u2 = −19.6 × 19.6 u2 = 19.62 u = 19.6 m/s Hence, initial velocity is 19.6 m/s Finding Time Taken (t) We have u, v and a Finding t using 1st equation of motion v = u + at 0 = 19.6 + (−9.8) t 9.8t = 19.6 t = 19.6/9.8 t = 2 seconds ∴ Time taken is 2 seconds Question A cricket ball is dropped from a height of 20 meters. Calculate the speed of the ball when it hits the ground. Calculate the time it takes to fall through this height (g = 10 m/s2) Since the ball is dropped, initial velocity = u = 0 m/s Distance travelled = s = 20 m And, Acceleration = a = g = 10 m/s2 (Acceleration is positive because stone falls downwards) Finding speed of the ball when it hits the ground. We need to find final velocity (v) We know u, s and a, Finding v using 3rd equation of motion v2 − u2 = 2as v2 − (0) 2 = 2 × 10 × 20 v2 = 20 × 20 v2 = 202 v = 20 m/s Speed with which ball hits ground is 20 m/s Finding the time it takes to fall through this height We need to find t We know u, v and a, Finding t using 1st equation of motion v = u + at 20 = 0 + (10) t 20 = 10t 10t = 20 t = 20/10 t = 2s Therefore, Time Taken is 2 seconds Question A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall (g = 9.8 m/s2) Given Initial velocity = u = 15 m/s At the highest point of motion, velocity becomes 0 Therefore Final velocity = v = 0 m/s And Acceleration = −g = −9.8 m/s2 (Acceleration is negative because ball is thrown upwards) We need to find the maximum height (s) We know u, v & a, finding s using 3rd equation of motion v2 − u2 = 2as 02 − 152 = 2 × (−9.8) × s –225 = –19.6 × s 225 = 19.6 × s 19.6 × s = 225 s = 225/19.6 s = 2250/196 = 11.47 m Max height is 11.47 m