A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.30 m above. The keys are caught 1.30 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 upward or downward
Can someone help me set up ?Or explain it please. Thank you!
ايوه صح كلامك يكبير بس نمره بي انت هابدها ولا اه
To solve this problem, we can use the equations of motion.
(a) Firstly, let's determine the initial velocity of the keys. The initial velocity (Vi) can be calculated using the equation:
Vi = (d - 0.5 * g * t^2) / t
where
d = displacement (the height of the window) = 4.30 m
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time taken to catch the keys = 1.30 s
Plugging in the values into the equation, we have:
Vi = (4.30 - 0.5 * 9.8 * (1.30)^2) / 1.30
Simplifying the equation, we get:
Vi = 12.298 m/s (upward)
So, the initial velocity of the keys is approximately 12.298 m/s upward.
(b) Now, let's determine the velocity of the keys just before they were caught. The final velocity (Vf) can be calculated using the equation:
Vf = Vi - g * t
where
Vi = initial velocity = 12.298 m/s (upward)
g = acceleration due to gravity = 9.8 m/s^2
t = time taken to catch the keys = 1.30 s
Plugging in the values into the equation, we have:
Vf = 12.298 - 9.8 * 1.30
Simplifying the equation, we get:
Vf = 0 m/s
So, the velocity of the keys just before they were caught is 0 m/s (neither upward nor downward).
To solve this problem, we can use the equations of motion for an object in free fall. The equations we need are:
1. h = ut + (1/2)gt^2 - This equation gives the height (h) of the object at a given time (t) when it is thrown upwards. Here, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time since it was thrown upwards.
2. v = u + gt - This equation gives the final velocity (v) of the object at a given time (t). Here, u is the initial velocity, g is the acceleration due to gravity, and t is the time since it was thrown upwards.
(a) To find the initial velocity with which the keys were thrown, we can use the first equation and solve for u. We know that the height h is 4.30 m, and the time t is 1.30 s. Using these values, we can rearrange the equation to solve for u:
h = ut + (1/2)gt^2
4.30 = u(1.30) + (1/2)(9.8)(1.30^2)
Simplifying this equation will give the initial velocity u. You can use a calculator or algebraic manipulation to find the value of u.
(b) To find the velocity of the keys just before they were caught, we can use the second equation. We know that the acceleration due to gravity is acting in the downward direction, so it will be negative. The final velocity will be zero because at the moment of catching, the keys are at rest.
v = u + gt
0 = u + (-9.8)(1.30)
Simplifying this equation will give the velocity u. If the value of u is positive, it means the keys were caught with an upward velocity. If the value of u is negative, it means the keys were caught with a downward velocity. Again, you can use a calculator or algebraic manipulation to find the value of u.
(a) 4.30 = Vo*t - (g/2) t^2
= Vo*1.3 - (1/2)*9.81*(1.3)^2
Solve for Vo.
1.3*Vo = 4.30 + 8.29 = 12.59 m
Vo = 9.68 m/s
(b) Use conservation of energy
Initial KE - Final KE = M g * 4.30 m
(Vo^2 - Vf^2)/2 = g * 4.30 m
Solve for the final velocity Vf. The keys could be on the way up or down.