Quick help with a physics problem?
A pair of glasses is dropped from a stadium 32 m high. A pen is dropped two seconds later. (disregard air resistance (g=-9.8)How far from the ground will the pen be when the glasses hit the ground?
i used the forumla; displacement=Vi(t)+ 1/2a(t squared) to start out i dont really know please helpppp!
Comments
Yes ... you've chosen the correct formula to use
Air resistance is being ignored ... so the time taken by the glasses and by the pen to drop the 32 m to the ground is the same
So first find the time it takes the glasses to fall to the ground:
Consider the vertical motion
They have given g = -9.8 m/s² ... so that means UP is the positive direction ...
s = -32 m ... [don't forget the negative b/c the displacement is down ... so negative]
a = g = -9.8 m/s²
v(i) = 0 m/s ... [b/c the glasses are just dropped]
t = ?
s = v(i)t + (1/2)at²
-32 = 0 +(1/2) *(-9.8)t²
-32 = -4.9t²
t = √(32/4.9) ... [negatives cancel]
t = 2.6 s ... [rejecting the negative possibility b/c time is positive]
so it takes the glasses 2.6 s to hit the ground
The pen is dropped 2 s after the glasses ... so the pen is falling for 2.6 - 2 = 0.6 s when the glasses hit the ground
now have to find the displacement of the pen during the first 0.6 s of its fall:
s = ? ... [it's down ... so expect a negative answer]
v(i) = 0 m/s ... [it's just dropped]
a = -g = -9.8 m/s²
t = 0.6 s
s = v(i)t + (1/2)at²
s = 0 + (1/2) * (-9.8) * 0.6²
s = -1.8 m ... [expecting negative answer ... b/c down is the negative direction]
so the distance the pen is from the ground when the glasses land = 32 - 1.8 = 30.2 m