A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
1 answer:
Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration
where
Vo = Initial speed =0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time
T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed
c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares
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(a) +2.60 m/s
The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:
- a horizontal uniform motion, at constant speed
- a vertical motion, at constant acceleration (acceleration of gravity, , downward)
In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):
(b) -17.2 m/s
The vertical component of the fish's velocity instead follows the equation:
where
is the initial vertical velocity, which is zero
is the acceleration of gravity
t is the time
Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:
where the negative sign indicates the direction (downward).
(c)
The horizontal component of the fish's velocity would increase
The vertical component of the fish's velocity would stay the same.
As we said from part (a) and (b):
- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too
- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.