During a tennis serve, a racket is given an angular acceleration of magnitude 155 rad/s2. at the top of the serve, the racket has an angular speed of 20.0 rad/s. if the distance between the top of the racket and the shoulder is 1.40 m, find the magnitude of the total acceleration of the top of the racket.

600.6 m/s^2

Explanation:

α = 155 rad/s^2

ω = 20 rad/s

r = 1.4 m

Tangential acceleration, aT = r x α = 1.4 x 155 = 217 m/s^2

Centripetal acceleration, ac = rω^2 = 1.4 x 20 x 20 = 560 m/s^2

The tangential acceleration and the centripetal acceleration both are perpendicular to each other. Let a be the resultant acceleration.

a^2 = aT^2 + ac^2

a^2 = 217^2 + 560^2

a = 600.6 m/s^2

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.