The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m
<span>H(t) = -16t^2 + vt + s
</span><span>Part A:
</span>Using the given data:
H(t)= -16*t² + 60*t + 82;
Part B:
Put H(t)=0
0<span>= -16*t² + 60*t + 82;</span>
Use the quadratic formula to find t.
See the attachment...'t' is replaced with 'x'.
Answer:
time of collision is
t = 0.395 s

so they will collide at height of 5.63 m from ground
Explanation:
initial speed of the ball when it is dropped down is

similarly initial speed of the object which is projected by spring is given as

now relative velocity of object with respect to ball

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m



Now the height attained by the object in the same time is given as



so they will collide at height of 5.63 m from ground
Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
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