K.E = 1/2*m*v^2 = 1/2(500)(3)^2 = 2250 J
m*g*h = 500(9.8)(30) = 147000 J
2250 + 147000 = 149250
Answer:
The maximum height the pebble reaches is approximately;
A. 6.4 m
Explanation:
The question is with regards to projectile motion of an object
The given parameters are;
The initial velocity of the pebble, u = 19 m/s
The angle the projectile path of the pebble makes with the horizontal, θ = 36°
The maximum height of a projectile,
, is given by the following equation;

Therefore, substituting the known values for the pebble, we have;

Therefore, the maximum height of the pebble projectile,
≈ 6.4 m.
Answer:
10.4 m/s
Explanation:
The problem can be solved by using the following SUVAT equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the diver in the problem, we have:
is the initial velocity (positive because it is upward)
is the acceleration of gravity (negative because it is downward)
By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.