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
You might be interested in
We can solve the problem by using the first law of thermodynamics:
where
is the change in internal energy of the system
is the heat absorbed by the system
is the work done by the system on the surrounding
In this problem, the work done by the system is
with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is
with a negative sign as well because it is released by the system.
Therefore, by using the initial equation, we find