Question

A hammer is thrown from the ground level of the Moon with a speed v0 = 10 m/s at an angle 45◦ above the horizontal. Assume that the acceleration due to gravity on the surface of the Moon is one sixth of the acceleration due to gravity on the surface of the Earth (take g = 10 m/s2 ).(i) Draw a coordinate system and sketch the hammer’s trajectory.(ii) Find the maximum height of flight of the hammer.(iii) Find the component of the hammer’s acceleration vector that is perpendicular to the hammer’s velocity vector at the highest point.(iv) What is the hammer’s velocity at the highest point?

Answer 1

Answer:

Explanation:

initial velocity, vo = 10 m/s

angle of projection, θ = 45°

acceleration due to gravity, g' = g/ 6 = 1.67 m/s^2

(i) diagram is attached.

(ii) The formula for the maximum height is given by

$H=\frac{u^{2}Sin^{2}\theta }{2g'}$

$H=\frac{10^{2}Sin^{2}45 }{2\times 1.67}$

H = 14.97 m

(iii) The acceleration of hammer is verticall downwards which is equal to g' = 1.67 m/s^2.

(iv) At heighest point, the hammer has horizontal component of velocity.

V = vo Cos 45 = 7.07 m/s

Answer 2

Answer:

Answer:

i. For the diagram of the hammer's trajectory, it is attached as an image to this solution

u = 10, θ=45°, g = 1/6 * 10 = 1.66667

ii. H = maximum height = $H = \frac{u^{2}sin^{2}\theta   }{2g}$

H = 10*10*(sin45)*(sin45)/2*1.66667

H = 14.9999m

iii. $v^{2}  = u^{2} + 2gH$

but v =  0 at max height

$v^{2} = 2gH\\=> g = \frac{v^{2} }{2H}$

g = $\frac{10^{2} }{2*14.9999}$

$g = 3.3334ms^{-2}$

iv.

The hammer's velocity at highest point is 0.

Explanation:

For the diagramatic illustration of the problem, it is attached as an image file.

The next thing to be done is to extract the available values for calculation. Now the initial velocity of the hammer is zero since it was thrown from the ground level of the moon

The acceleration due to gravity used in calculation was that of the moon which according to the question is 1/6 that of the moon.

Calculating the maximum height of the hammer just involved substituting the right values in the formula for a projectile's maximum height

To calculate the hammer's acceleration at the maximum height, we need to remember that the velocity of an object at maximum height is always zero. This is because at this instant, the velocity is reduced by the acceleration due to gravity before direction of the object changes. Now, we make use of the formula: $v^{2} = u^{2} + 2gH$. Now, we substitute v=0 because at maximum height, the velocity is 0. Our aim now is to find g. we substitute the values and remember we calculated the value of the maximum height earlier whose value we will be using in the calculation of g

The answer is zero. This is because at this instant, the velocity is reduced by the acceleration due to gravity before direction of the object changes.