Answer:
31.905 ft/s²
Explanation:
Given that
Mass of the pilot, m = 120 lb
Weight of the pilot, w = 119 lbf
Acceleration due to gravity, g = 32.05 ft/s²
Local acceleration of gravity of found by using the relation
Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)
119 = 120 * a/32. 174
119 * 32.174 = 120a
a = 3828.706 / 120
a = 31.905 ft/s²
Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²
Answer:
<h3> 1.40625m/s²</h3>
Explanation:
Using the equation of motion expressed as v = u+gt where;
v is the final velocity of the ball
u is the initial velocity
g is the acceleration due to gravity
t is the time taken
Given
u = 9m/s
v = 0m/s
t = 6.4s
Required
acceleration due to gravity g
Since the rock is thrown up, g will be a negative value.
v = u+(-g)t
0 = 9-6.4g
-9 = -6.4g
6.4g = 9
divide both sides by 6.4
6.4g/6.4 = 9/6.4
g = 1.40625m/s²
Hence the acceleration due to gravity on the planet is 1.40625m/s²
Answer:
Explanation:
If a baseball is hit into the air with a velocity of 27 m/s, we want to determine the maximum height of the ball. Using the projecile formula;
Max height H = u²/2g
u is the initial velocity of the body = 27m/s
g is the acceleration due to gravity = 9.81m/s²
H = 27²/2(9.81)
H = 729/19.62
H = 37.16m
Hence the ball went 37.16m high
The child on a swing can be modeled as a simple pendulum. The period of a simple pendulum is given by