Answer:
a) The velocity of the rock after 1 s is 14.28 m/s.
b) The velocity of the rock after "a" seconds is 18 m/s - 3.72 m/s² · a
c) The rock will hit the ground after 9.7 s.
d) The velocity with which the rock hits the surface is -18.1 m/s
Explanation:
Hi there!
Let´s write the equation:
H(t) = 18 m/s · t - 1.86 m/s² · t²
The velocity is the variaiton of the height over time, in other words, it is the derivative of the height function with respect to time. Then:
v(t) = dH/dt = 18 m/s - 2 · 1.86 m/s² · t
v(t) = 18 m/s - 3.72 m/s² · t
a) When t = 1 s:
v(1 s) = 18 m/s - 3.72 m/s² · 1 s
v(1 s) = 14.28 m/s
b) When t = a:
v(a) = 18 m/s - 3.72 m/s² · a
c) When the rock hits the surface, the height of the rock is 0:
H(t) = 18 m/s · t - 1.86 m/s² · t²
0 = 18 m/s · t - 1.86 m/s² · t²
0 = t(18 m/s - 1.86 m/s² · t)
t = 0
and
18 m/s - 1.86 m/s² · t = 0
t = -18 m/s /- 1.86 m/s²
t = 9.7 s
d) Let´s use the equation of velocity to find the velocity at t = 9.7 s when the rock hits the ground.
v(t) = 18 m/s - 3.72 m/s² · t
v(9.7 s) = 18 m/s - 3.72 m/s² · 9.7 s
v(9.7 s) = -18.1 m/s
