with what minimal horizontal velocity must the cliff divers leave the top of the cliff is they are to miss the rock?
1 answer:
14.8m/s is minimal horizontal velocity must the cliff divers leave the top of the cliff is they are to miss the rock.
Ledge divers in Acapulco leap into the water from a 36-meter-high cliff. A rock protrudes 10.4 meters horizontally from the water, level. 9.8 m/s2 is the acceleration of gravity. If the cliff divers are to miss the rock, they must depart from the top of the cliff with a minimal horizontal velocity.
from above statements, we assume that
there is no initial vertical velocity, then we calculate the time it takes to reach the water, then divide the horizontal distance by that time to get the minimum required horizontal speed.
h=gt^2/2
h= 36 m
t =
put h = 36 m g = 9.8 m/s^2
t=2.7 second
Then the minimum required horizontal velocity is
v = d/t
by putting value
v= 10.4 m/2.7 s
v=14.8m/s
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Answer: 6700m/s^2 E
Explanation:
Given the following :
Velocity of ball before being hit by a racquet = initial velocity(u) = 55m/s W
Velocity of the ball after being hit = final Velocity (v) = 45m/s E
Time of contact between ball and racquet (t) = 1.5 × 10^-2
The acceleration of a body is the change in Velocity of a body with time. It given by:
Acceleration (a) = [final Velocity(v) - initial velocity(u)] / time(t)
The westward direction is towards the left and thus Velocity takes a negative value, similarly, if it's towards the right, Velocity takes a positive value.
Therefore,
a = [45 - (-55)] / 1.5×10^-2
a = [45 + 55] / 0.015
a = 100 / 0.015
a = 6666.6666 m/s^2
a = 6700m/s^2 E
Since the value of acceleration is positive, the direction is towards the East (acceleration is in the direction of the ball's final Velocity).
