A ball is dropped from 10.0m. How long does it take to hit the ground?
Approximately 1.428 seconds. The formula for the duration of a fall based on height (h) and acceleration due to gravity (g) is:[math]t = \sqrt(2 h/g)[/math]
A ball is dropped from the height of 5m. How long does it take for the ball to hit the ground?
The motion is that of "constant acceleration" with no initial velocity. Applicable formula: D = 1/2 gt² (where g = 9.8 m/s²) solving for "t": D = 4.9 t² t² = D/4.9 = 5/4.9 = 1.02 t = √1.02 = 1.01 sec or approx 1 second (ANS)
A ball is dropped from a height of s=400-16t^2 meters. How long til it hits the ground?
How long does it take the ball to reach the ground? when height s = 0 0=400-16t^2 16t^2 = 400 t^2 = 400/16 = 25 t = + - 5 but u cant have negative time so the answer when it hit s the ground is 5 sec after it is dropped.
A ball is dropped from the height 1.9m how long does it take to hit the ground?
your answer is correct. you have displacement s=1.9m initial velocity u=0m/s acceleration=a=g=9.8m/s we have s=ut + (a/2)t^2 therefore t=0.62 s
Suppose you drop a 1-kg rock from a height of 5m above the ground. (the rock stops upon impact.)?
Answers: (1) Kinetic energy pre-impact = 49.1 J (2) Velocity pre-impact = 9.90 m/s (3) Momentum pre-impact= -9.90 m/s (4) Both kinetic energy and momentum are zero because the rock is stationary (5) See below for full explanation. With this information, you cannot determine the force, as force is currently infinite. I would do... Given: m = 1 kg d = 5 m g = 9.81 m/s^2 Vi = 0.0 m/s Find velocity before impact through uniform acceleration Vf = SQRT { Vi^2 + [2gd] } Vf = SQRT { (0.0 m/s)^2 + [2 * (9.81 m/s^2) * (5 m) ] Vf = SQRT { 98.1 m^2/s^2 } Vf = 9.90 m/s Now you can find the kinetic energy of the rock KE = 0.5 * m * v^2 KE = 0.5 * (1 kg) * (9.90 m/s)^2 KE = (0.5 kg) * (98.1 m^2/s^2) KE = 49.1 J Momentum of the rock before impact M = m * v M = (1 kg) * (-9.90 m/s) M = -9.90 m/s KE and M after impact, assuming instantaneous stop KE = 0 M = 0 Because v = 0 m/s, which multiplies to 0 To determine force from the ground. You would use Newton's 2nd Law: F = m * a However, a is acceleration which means change in velocity over time. a = [Vf - Vi] / t But as time is instantaneous stop, there is no t value other than 0. Everybody knows that when you divide by zero, you gen an infinite result. Obviously, the force cannot be infinite. Therefore, with this information, you cannot determine the force.
A ball is dropped from 3 meters. how long does it take to reach the ground?
You have the initial velocity (0m/s), the displacement (3m) and the acceleration (9.81m/s^2), and you wish to find the time. Therefore choose the constant acceleration formula that links these four terms together: s = ut + 0.5at^2 3 = 0 + 0.5*9.81*t^2 3 = 4.905t^2 t^2 = 3 / 4.905 t^2 = 0.61162 t = 0.78 s to two significant figures. Now you wish to find the final velocity, now that you have the time, you can pick from two formulae, but for best accuracy I'd choose the one that does not rely on the time calculated (after all, that answer has just been rounded off). v^2 = u^2 + 2as v^2 = 0 + 2*9.81*3 v^2 = 58.86 v = 7.7 m/s to two significant figures.