Question

An electronic line judge camera captures the impact of a 57.0-g tennis ball traveling at 32.2 m/s with the side line of a tennis court. The ball rebounds with a speed of 21.6 m/s and is seen to be in contact with the ground for 3.94 ms. What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground

Answer 1

Answer:

average acceleration is 1.365 × $10^{4}$ m/s²

Explanation:

given data

initila speed  u = -32.2 m/s

final speed v = 21.6 m/s

time taken t = 0.00394 s

solution

we get here average acceleration that will be express as

v = u + at    ..........................1

put here value and we get

21.6 = -32.2 + a × 0.00394

solve it we get

a = 1.365 × $10^{4}$ m/s²

so average acceleration is 1.365 × $10^{4}$ m/s²