# Answer to Question #69094 in Mechanics | Relativity for Tianna

Question #69094

A ball is thrown vertically upward. The quantity which stays constant is a) displacement b)acceleration c)velocity d)speed e)momentum.

2) A ball is thrown from the ground into the air. At a height of 9.6m, the velocity is observed to be v= 8.0i + 6.5j in m/s. Find the maximum height the ball can reach.

2) A ball is thrown from the ground into the air. At a height of 9.6m, the velocity is observed to be v= 8.0i + 6.5j in m/s. Find the maximum height the ball can reach.

Expert's answer

The acceleration stays constant or the answer b)

The second part:

Horizontal component of velocity is 8m/s

The velocity at 9.6m

v = sqrt(8^2 + 6.5^2)

v = 10.3 m/s

The whole energy at 9.6m is

E = 0.5mv^2 + mgh

E = m(0.5*10.3^2 + 9.8*9.6)

E = 147.125m

Energy at initial position is equal to the Energy at final position

147.125m = mgh + 0.5mv^2

Maximum height v = horizontal compnent of velocity

147.125m = m*9.8*H + 0.5*m*64

147.125 = 9.8H + 32

H = 11.8 m

The second part:

Horizontal component of velocity is 8m/s

The velocity at 9.6m

v = sqrt(8^2 + 6.5^2)

v = 10.3 m/s

The whole energy at 9.6m is

E = 0.5mv^2 + mgh

E = m(0.5*10.3^2 + 9.8*9.6)

E = 147.125m

Energy at initial position is equal to the Energy at final position

147.125m = mgh + 0.5mv^2

Maximum height v = horizontal compnent of velocity

147.125m = m*9.8*H + 0.5*m*64

147.125 = 9.8H + 32

H = 11.8 m

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