Question

If a 0.5 kg ball is thrown up with 250 J of kinetic energy, how high will it go?

(HINT: At the ball's highest point it has stopped.)
KE = 1/2mv2
G = 9.8 m/s2

Answer 1

51.02m

Step-by-step explanation:

KE = 1/2mv2

Where k.e = 250J

mass = 0.5 kg

g = 9.8 m/s2

250= 1/2×0.5×v^2

250= 0.5×0.5×v^2

250= 0.25v^2

v^2 = 250/0.25

v^2 = 1000

v =√1000

v = 31.62m/s

v^2= u^2-2gh........... (1)

Since the object will stop at it highest point, hence it final velocity there will be zero and since it is moving up against the gravity g= -9.8m/s^2. That was why the formula in equation 1 has a negative sign

From h = u^2/2g

Where v = 31.62m/s

g = 9.8m/s^2

H = (31.62m/s)^2/9.8×2

H= 1000/19.6

= 51.02m

Hence the height of it travelling will be

51.02m