A ball is thrown straight upward. When it has reached the highest point in its motion, and is momentarily stopped, its accelerat
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Given
v = 343 m/s
ac = 5g
ac = 5*9.8 m/s^2
ac = 49 m/s^2
where,
v: velocity
ac = centripetal aceleration
Procedure
We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.
Formula
The minimum radius not to exceed the centripetal acceleration is 2401 m.
(a) 3.35 s
The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration
a = g = -9.8 m/s^2
towards the ground.
The initial height of the projectile is
h = 55.0 m
The vertical position of the projectile at time t is
By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:
(b) 78.4 m
The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.
The horizontal motion is a uniform motion with constant velocity -
The horizontal velocity of the projectile is
the time it takes the projectile to reach the ground is
t = 3.35 s
So, the horizontal distance covered by the projectile is
(c) 23.4 m/s, -32.8 m/s
The motion of the projectile consists of two independent motions:
- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to
so this value is also the value of the horizontal velocity just before the projectile reaches the ground.
- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by
where
is the initial vertical velocity
Using
a = g = -9.8 m/s^2
and
t = 3.35 s
We find the vertical velocity of the projectile just before reaching the ground
and the negative sign means it points downward.