Answer:
Vf = 14.7 m/s
Explanation:
Vf² = Vi² + 2 * a * Δy
given:
a = 9.81 m/s²
Δy = 11m
Vi = 0 when upon release
Vf² = 0 + 2 (9.81) 11
Vf = 14.7 m/s
The velocity of the ball when it hits the ground will be 14.7 m/s.
The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
given:
a(gravitational acceleration) = 9.81 m/s²
s (distance)= 11m
v is final velocity
u is the initial velocity
From Newton's second equation of motion;
[tex]\rm v^2 = u^2+2as \\\\ v^2=2 \times 9.81 \times 11 \\\\ v= 14.7 \ m/sec[/tex]
Hence, the velocity of the ball when it hits the ground will be 14.7 m/s.
To learn more about the velocity refer to the link ;
https://brainly.com/question/862972
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