Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.
N/m
Explanation:
Answer:
0.34s, 8.5m,31.89m
Explanation:
The above motion defines a projectile motion.
Now the athletes lands on a cliff 30° to the horizontal this means the velocity at that point would be 25m/s cos30°
Now from Newton's law of motion.
The body would be decelerating so,
V = u - gt
Where u is initial velocity and v is final velocity. g is acceleration of free fall due to gravity.
Hence,
V-U/ -g = t
Hence 25cos30 - 25/ -9.8 = 0.34s.
2.Now the length of the jump is defined as the total horizontal distance which is marked off by the horizontal velocity and time taken for take off and landing.
Hence Distance,S = u × t
25 ×0.34 =8.5m.
3. The maximum height is defined that at that point the Final velocity is 0m/s
Now the initial velocity is 25m/s
From Newton's law that;
V2= U2 -2gH; where U and V are initial and final velocity and H is height.
Hence H = V2-U2/-2g
=(0)^2- (25)^2/ -2×9.8
= -625/-19.6 =31.89m
