Question

A stone is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, the height of the building is 45.0 m. How long is the stone in flight?

Answer 1

Answer:

t =  4.218 s  :  stone flight time

Explanation:

The stone describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 20.0 ° m/s  , at an angle  α₀=30.0° above the horizontal

y₀ = 45.0 m

g= 9.8 m/s²

Calculation of the time it takes for the stone to hit the ground

v₀y =  v₀*sinα = (20 m/s)*sin(30°) = 10 m/s

We replace data in the equation (2)

y= y₀ + (v₀y)*t - (1/2)*gt²

0=  45 + (10)*(t ) - (1/2)*(9.8)(t )²

(4.9)(t )² - (10)(t ) -45 = 0  

We solve the quadratic equation:

t₁ = 4.218 s

t₁ = -2.177 s

Time cannot be negative therefore t₁ = 4.218 s is the time that the stone remains in the airt.

t =  4.218 s  :  stone flight time