Answer:
v = 17.71 m / s
Explanation:
We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity
v² = v₀² - 2 g (y -y₀)
v² = 0 - 2g (y -y₀)
when it hits the stone the height is zero and part of the height of the seagull I
v² = 2g y₀
v = Ra (2g i)
let's calculate
v =√ (2 9.8 16)
v = 17.71 m / s
Water is treated by purifying it by adding slaked lime or potash alum
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are


where
is the initial velocity.
(a).
When the projectile hits the 50m mark,
; therefore,

solving for
we get:

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

which gives

(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

the vertical component of the velocity is

which gives a speed
of


Answer:
the extension would be less the new extension might be 3 cm
Explanation: