Question

In the fastest measured tennis serve, the ball left the racquet at 73.14 m/s. A serve tennis ball is typically in contact with the racquet for 30.0 ms and starts from rest. Assume constant acceleration.(a) what was the ball's acceleration during this serve??(b) how far did the ball travel during the serve???

Answer 1

Answer:

a)  the acceleration is a= 2438 m/s²

b) the distance travelled during serve is d = 1.0971 m

Explanation:

a) since

v = vo + a*t ,

where v= velocity at time t , vo= velocity at time t=0 and a= acceleration

,then

a= (v-vo)/t

replacing values

a= (v-vo)/t = (73.14 m/s - 0 m/s)/( 30* 10⁻³ s) = 2438 m/s²

b) the distance travelled d is

v² = vo² + 2*a*d  

then

d = (v² - vo²) /(2*a) = (73.14 m/s)² - 0²)/(2*2438 m/s²)= 1.0971 m

Answer 2

a)  the acceleration is a= 2438 m/s²

b) the distance travelled during serve is d = 1.0971 m

What is acceleration?

Acceleration represents the rate at which velocity should be changed with time, with respect to both speed and direction. Since acceleration contains both a magnitude and a direction, it is a vector quantity.

Calculation of acceleration & distance:

a) since

$v = vo + a\times t$

Here

v= velocity at time t ,

vo= velocity at time t=0

and a= acceleration

Now

$a= (v-vo)\div t\\\\ =(73.14 m/s - 0 m/s)/( 30\times 10^{-3} s)$

= 2438 m/s²

b) Now the distance traveled d is

$v^2 = vo^2 + 2\times a\times d \\\\d = (v^2 - vo^2) \div (2\timesa) \\\\=(73.14 m/s)^2 - 0^2)\div (2\times 2438 m/s^2)$

= 1.0971 m

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