An object is thrown downwards with a speed of 10m/s from a cliff that is 120m above ground. If air resistence is negligible, determine the time taken to reach the ground.
-
x = ut + (1/2)at^2
x = displacement = - 120 m
u = initial velocity = - 10 m/s
a = acceleration = - 9.8 m/s^2
t = time = ?
- 120 = - 10t - 4.9t^2
4.9t^2 + 10t - 120 = 0
use the quadratic formula to solve for t
t = (-b +/- (b^2 - 4ac)^(1/2))/ (2a)
t = 4.03
t = - 6.07
therefore, it took 4.03 seconds to reach the ground [event occured after t = 0]
x = displacement = - 120 m
u = initial velocity = - 10 m/s
a = acceleration = - 9.8 m/s^2
t = time = ?
- 120 = - 10t - 4.9t^2
4.9t^2 + 10t - 120 = 0
use the quadratic formula to solve for t
t = (-b +/- (b^2 - 4ac)^(1/2))/ (2a)
t = 4.03
t = - 6.07
therefore, it took 4.03 seconds to reach the ground [event occured after t = 0]