A grasshopper jumps a horizontal distance of 1.50 m from rest, with an initial velocity at a 43.0° angle with respect to the hor
2 answers:
Answer:
(a)v₀ =3.84 m/s
(b) h= 0.35 m
Explanation:
The grasshopper 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)
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
g: acceleration due to gravity in m/s²
Data
g = 9.8 m/s²
v₀ , at an angle α=43.0° above the horizontal
xi=0
x= 1.50 m
y₀=0
x-y components of v₀
v₀x = v₀cos43.0° = vx
v₀y = v₀sin43.0°
Calculation of the time it takes for the grasshopperl to toch the ground
We replace data in the equation (1)
x = 0 + vx*t = (v₀cos43.0°)*t
t= x/(v₀cos43.0°)
t= 1,5/(v₀cos43.0°) Equation (3)
We replace t= 1,5/(v₀cos43°),y₀=0, y = 0,:when the grasshopper touches the ground in the equation (2)
y=(v₀y)*t - (1/2)*g*t²
0=(v₀sin43°)*(1,5/(v₀cos43°) - (1/2)*(9.8)*1,5²/(v₀cos43°)²
0= 1,5*tan43° - 11.025/(v₀cos43°)²
11.025/(v₀cos43°)² = 1,5*tan43°
11.025 = (1,5*tan43°)*(v₀cos43°)²
(v₀cos43°)² = (11.025 )/ 1,5*(tan43°)
v₀² = (11.025 )/ (1,5*tan43°)(cos43°)²
v₀² = 14.73
v₀ = 3.84 m/s
Flight time
in the Equation (3):
t = 1,5/(v₀cos43.0°) = 1,5/(3.84*cos43°)
t = 0.534 s
(b) Maximum height reached (h)
The time to reach maximum height (h) is half the time of flight time :
th=0.534 s /2 = 0.267 s
We apply the Equation (2)
y= (v₀y)*t - (1/2)*g*t²
h= ( 3.84*sin 43°)*( 0.267) - (1/2)*(9.8)*( 0.267)²
h= 0.35 m
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