Agrasshopper jumps a horizontal distance of 1.50 m from rest, with an initial velocity at a 43.0° angle with respect to the horizontal.
(a) find the initial speed of the grasshopper.
m/s
(b) find the maximum height reached.
m

(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

the area between the graph and the t axis

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