A 430 g soccer ball lying on the ground is kicked with a force of 1000 N.?

Question

If the kick lasts 0.01 s, the ball flies off with what speed?

A. 0.023 m/s
B. 2.3 m/s
C. 0.23 m/s
D. 23 m/s

Answer 1

ft = mv
=> 1000 * 0.01 = 0.43 * v
=> v = 23 m/s

Answer 2

E = F x T

E = 1000 x 0.01 = 10 joules

kinetic energy = 1/2 x M x V^2

Use 430 g = .43 kg

10 = 1/2 x .43 x V^2

V = sqr (20/.43) = 6.81 m /sec ?????????????

Answer 3

First, we know by Isaac Newton's second law of motion that:
a = F/m, and the SI unit of F is Newtons, and of m is kg
so, 430 g = 0.43 kg
Then, we do:

a = 1000/0.43 = 2325.58 m/s^2

Then, because
v = at,
v = 0.01 * 2325.58 = 23.25 approximately 23

D

Answer 4

F=ma --> a=F/m

a= 1000/0.43
=2326 m/s^2

v=0.5 a*t^2
= 0.5 * 2326 * 0.01 * 0.01
= 0.116 m/s

Answer: none of the above

Answer 5

F=ma
1000=0.43((v-u/t)) where u, initial velocity of ball=0

1000=0.43 (v/0.01)
1=0.43v
v=2.33m/s

Answer 6

i agree with FedUp