Respuesta :
Answer:
a) 3.675 m
b) 3.67m
Explanation:
We are given acceleration due to gravity on earth =[tex]9.8ms^-2[/tex]
And on planet given = [tex]2.0ms^-2[/tex]
A) Since the maximum jump height is given by the formula
[tex]\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}[/tex]
Where H = max jump height,
v0 = velocity of jump,
Ø = angle of jump and
g = acceleration due to gravity
Considering velocity and angle in both cases
[tex]\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}[/tex]
Where H1 = jump height on given planet,
H2 = jump height on earth = 0.75m (given)
g1 = 2.0[tex]ms^-2[/tex] and
g2 = 9.8[tex]ms^-2[/tex]
Substituting these values we get H1 = 3.675m which is the required answer
B) Formula to find height of ball thrown is given by
[tex]\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}[/tex]
which is due to projectile motion of ball
Now h = max height,
v0 = initial velocity = 0,
t = time of motion,
a = acceleration = g = acceleration due to gravity
Considering t = same on both places we can write
[tex]\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}[/tex]
where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations
substituting h2 = 18m, g1 = 2.0[tex]ms^-2[/tex] and g2 = 9.8[tex]ms^-2[/tex]
We get h1 = 3.67m which is the required height
