Basic Physics

# Determine the maximum height of projectile motion

1. A kicked football leaves the ground at an angle θ = 60o with the horizontal has an initial speed of 10 m/s. Calculate the maximum height! Acceleration of gravity is 10 m/s2.

Known :

Angle (θ) = 60o

Initial speed (vo) = 10 m/s

Wanted : Maximum height (h)

Solution :

Vertical component of initial velocity :

sin 60o = voy / vo

voy = vo sin 60o = (10)(sin 60o) = (10)(0.53) = 53 m/s

Choose upward direction as positive and downward direction as negative.

Known :

Acceleration of gravity (g) = -10 m/s2 (negative downward)

Vertical component of initial velocity (voy) = +53 m/s (positive upward)

Final velocity at the maximum height (vty) = 0

Wanted : Maximum height (h)

Solution :

vt2 = vo2 + 2 g h

02 = (53)2 + 2 (-10) h

0 = 25(3) – 20 h

0 = 75 – 20 h

75 = 20 h

h = 75 / 20

h = 3.75 meter

The maximum height is 3.75 meter.

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2. A body is projected upward at angle of 30o with the horizontal from a building 20 meter high. It’s initial speed is 4 m/s. Calculate the maximum height! Acceleration of gravity is 10 m/s2.

Known :

Angle (θ) = 30o

Initial height (h) = 20 meter

o) = 4 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : The maximum height (h)

Solution :

Vertical component of initial velocity :

sin 30o = voy / vo

voy = vo sin 30o = (4)(sin 30o) = (4)(0.5) = 2 m/s

Choose upward direction as positive and downward direction as negative.

Known :

Acceleration of gravity (g) = -10 m/s2 (negative downward)

Vertical component of initial velocity (voy) = +2 m/s (positive upward)

Final velocity at maximum height (vty) = 0

Wanted : The maximum height

Solution :

The maximum height :

vt2 = vo2 + 2 g h

02 = 22 + 2 (-10) h

0 = 4 – 20 h

4 = 20 h

h = 4 / 20

h = 0.2 meter

The maximum height is 0.2 meter + 20 meter = 20.2 meter.

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