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A 50 kg figure skater is thrown upward by her partner with a vertical velocity o

ID: 1335381 • Letter: A

Question

A 50 kg figure skater is thrown upward by her partner with a vertical velocity of 2 m/s. He releases her a height of 2.5 m above the ground. he misses when trying to catch her. Compute the following: A. Velocity when she is 2.5m above the ground on the way down B. The velocity at impact with the ground C. The peak height D. The time to hit the ground A 50 kg figure skater is thrown upward by her partner with a vertical velocity of 2 m/s. He releases her a height of 2.5 m above the ground. he misses when trying to catch her. Compute the following: A. Velocity when she is 2.5m above the ground on the way down B. The velocity at impact with the ground C. The peak height D. The time to hit the ground A 50 kg figure skater is thrown upward by her partner with a vertical velocity of 2 m/s. He releases her a height of 2.5 m above the ground. he misses when trying to catch her. Compute the following: A. Velocity when she is 2.5m above the ground on the way down B. The velocity at impact with the ground C. The peak height D. The time to hit the ground

Explanation / Answer

The skater is released from a height of 2.5m above ground.

From law of conservation of energy, kinetic energy of skater is equal to the potential energy at the peak height. Let u be the intial velocity given to be 2m/s, m = 50 kg and let h' be the peak height above 2.5m above ground. Take g = 10m/s2

Therefore, (1/2)mu2= mgh' => h'= 0.2m

Therefore, maximum height = 0.2+2.5 = 2.7m

A. On the way down, the skater experiences a free fall. From 3rd equation of motion, v2=u2+2gh'

=> v2=0+2*10*0.2 => v = 2m/s

B. Velocity at impact, (1/2)*m*v2 = m*g*(2.7) = 7.348m/s (conservation of energy)

C. Peak height is 2.7m as calculated above.

D. Time to reach maximum height, t1:

v = u-gt1 => 0 = 2-10.t1 => t1 = 0.2 seconds

Time to reach ground from maximum height, t2:

v = u+gt2 => 7.348 = 0+ 10t2 => t2 = 0.7348 seconds.

Hence time to hit ground is t1+t2 = 0.2+0.7348 = 0.9348 seconds.

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