hi guys, i had the following question but i keep on getting it wrong, should i use -9.8 or 9.8 m/s2 for the acceleration? im confused and would really appreciate some help.
A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.70 m above. The keys are caught 1.60 s later by the sister's outstretched hand.
(a) With what initial velocity were the keys thrown?
m/s upward
(b) What was the velocity of the keys just before they were caught?
m/s is it upwards or downwards?
thanks to anyone who can help me
the sign mean direction. Gravity is downward, so if you have + as the upward direction, then gravity is downward and negative.
vf^2=vi^2+ 2acceleration*height
when the keys are caught, one wants vf to be zero.
0=vi^2+ 2 (-9.8)(+3.70)
not the plus signs upward on the height, but negative for gravity.
Solve for vi.
Just before they were caught? They were at the top, going up, near zero.
g = -9.8 m/s^2 if the positive direction fot the vertical axis is defined as up.
g = +9.8 m/s^2 if the positive direction fot the vertical axis is defined as down.
You can do it either way.
(a) Assume that the keys are caught while moving up. If Y is measured upwards from the lower sister's hand,
Y = V*1.6 - (1/2) g *(1.6)^2 = 3.7 m
1.6 V = 3.7 + 12.5
Solve for V, the initial velocity.
V = +10.1 m/s
(b) When caught, v = V - gt
v = -5.6 m/s
The keys are headed back down when caught.
To solve this problem, we first need to understand the motion of the keys. Since the keys are thrown vertically upward and then caught, we can assume the acceleration due to gravity (-9.8 m/s^2) is acting in the downward direction throughout the motion.
(a) To find the initial velocity of the keys, we can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s, as the keys come to a stop at their highest point)
u = initial velocity (what we need to find)
a = acceleration (-9.8 m/s^2)
t = time (1.60 s)
Rearranging the equation, we get:
u = v - at
Substituting the known values:
u = 0 - (-9.8 * 1.60)
u = 15.68 m/s upward
Therefore, the initial velocity with which the keys were thrown was 15.68 m/s upward.
(b) To find the velocity of the keys just before they were caught, we can use the same kinematic equation. However, this time we need to find the final velocity.
Using the equation:
v = u + at
Substituting the known values:
u = 15.68 m/s upward (from part a)
a = -9.8 m/s^2 (acceleration due to gravity)
t = 1.60 s
v = 15.68 - (9.8 * 1.60)
v = 0 m/s
Therefore, the velocity of the keys just before they were caught was 0 m/s. Note that the velocity is 0 because at the highest point of its trajectory, the keys come to a stop before falling back down.
In summary, the initial velocity of the keys was 15.68 m/s upward, and the velocity just before they were caught was 0 m/s.