A stone is thrown vertically upward with a speed of 35.0 m/s a) Howfast sit moving when it reaches a height of 13.0m How much time is required to reach this height? Wht s the maximum height it will reach? Explain why there are two answers for part 3

a)

v = 31.15 m/s

t = 0.393 sec

h = 62.5 m

Explanation:

a)

v₀ = initially speed of the stone = 35.0 m/s

v = final speed of the stone = ?

y = vertical displacement of the stone = 13.0 m

a = acceleration due to gravity = - 9.8 m/s²

using the equation

v² = v₀² + 2 a y

v² = 35² + 2 ( - 9.8) (13.0)

v = 31.15 m/s

t = time taken

using the equation

v = v₀ + a t

31.15 = 35 + ( - 9.8) t

t = 0.393 sec

h = maximum height

v' = final speed at the maximum height = 0 m/s

using the equation

v'² = v₀² + 2 a h

0² = 35² + 2 ( - 9.8) h

h = 62.5 m