solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
The volume of the block of wood is given by length × width ×height
= 2.75 × 4.80 × 7.5
= 99 cm³
Density is given by mass/volume
Thus = 84.0 g/ 99 cm³
= 0.848 g/cm³
Hence; since the block is less dense than water (1 g/cm³) it will float
Solution:
Given:
height of geyser, h = 50 m
speed of water at ground can be given by third eqn of motion with v = 0 m/s:

putting v = 0 m/s in the above eqn, we get:
u =
(1)
Now, pressure is given by:
(2)
where,
= density of water = 1000 kg/
g = 9.8 m/
Using eqn (1) and (2):

⇒ 
=
= 490000 Pa
= 101325 Pa
For absolute pressure, p:
p =
+
p = 490000 - 101325
p = 388675 Pa
Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa