A rock is thrown 0.8 meters into the air. how fast was it thrown?
1 answer:
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Answer:
t = 2.77 s
Explanation:
The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:
Equation of movement of the ball in the X axis
X = v₀x*t Equation (1)
Equation of movement of the ball in the Y axis
Y = y₀+ v₀y*t -½ g*t² Equation (2)
Where
X : horizontal position in meters (m)
Y : vertical position in meters (m)
y₀ : initial vertical position in meters (m)
v₀x : X-initial speed in m/s
v₀y : Y-initial speed in m/s
g: acceleration due to gravity in m/s²
t : time to position (X,Y)
Data
y₀ = 37.5 m
v₀y = 0
g = 9.8 m/s²
Problem development
The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0
We apply the Equation (2):
Y = y₀+ (v₀y)*t - (½) g*t²
0 = 37.5 +(0)*t- (1/2)*(g)*t²
0 = 37.5 - (1/2)*(9.8)*t²
(1/2)*(9.8)*t² = 37.5
t² = (2)(37.5)/(9.8)
t = 2.77 s