The relevant equation we can use in this problem is:
h = v0 t + 0.5 g t^2
where h is height, v0 is initial velocity, t is time, g is
gravity
Since it was stated that the rock was drop, so it was free
fall and v0 = 0, therefore:
h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2
<span>h = 117.77 m</span>
Answer:
Explanation:
Given
Initial velocity u = 200m/s
Final velocity = 4m/s
Distance S = 4000m
Required
Acceleration
Substitute the given parameters into the formula
v² = u²+2as
4² = 200²+2a(4000)
16 = 40000+8000a
8000a = 16-40000
8000a = -39,984
a = - 39,984/8000
a = -4.998m/s²
Hence the acceleration is -4.998m/s²
V^2=u^2+2as
V=0
a =-u^2/2s
a=[4]^2/2[4]
a=-2m/s^2
Answer:
its a solid but can flow
Explanation:
those answers to choose from are wrong
Answer:
Explanation:
compressions and rarefactions
