An object is thrown upwards at an initial speed of 19.6m/s from a height of 55m. How long will it take for the object to reach the ground.(from the time it was thrown)
Solve this equation for t:
H(t) = 55 + 19.6 t -4.9 t^2 = 0
Take the positive root. (There will be two solutions.)
Drwls
Could it also be solved by using
v=u+at for the upward throw;then another motion equation for the upward distance; added to the fall time (sqrt 2h/g) all the way to the bottom?
Charlie
Yes, that would work also.
To find the time it takes for the object to reach the ground, we can use the kinematic equation that relates the initial velocity, final velocity, time, and displacement of an object in free fall. The equation is:
s = ut + (1/2)gt^2
Where:
s = displacement (final position minus initial position)
u = initial velocity
t = time
g = acceleration due to gravity (approximately -9.8 m/s^2, negative because it acts in the opposite direction of the object's motion)
In this case, the object is thrown upwards, so we'll use a negative value for the initial velocity (u).
Given:
u = -19.6 m/s
s = -55 m
g = -9.8 m/s^2
Plugging in these values, the equation becomes:
-55 = -19.6t - (1/2)(9.8)t^2
Simplifying further:
0 = -9.8t^2 - 19.6t - 55
Since this is a quadratic equation, we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = coefficient of t^2 term (-9.8 in this case)
b = coefficient of t term (-19.6 in this case)
c = constant term (-55 in this case)
Substituting the values:
t = (-(-19.6) ± √((-19.6)^2 - 4(-9.8)(-55))) / (2(-9.8))
Simplifying further:
t = (19.6 ± √(384.16 - 2156)) / (-19.6)
t = (19.6 ± √(-1771.84)) / (-19.6)
Since the square root of a negative number is not a real value, this means that the object will never reach the ground.