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.
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
2 answers
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.
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.