An object of mass 500 g is projected in the air at an angle of 30 deg with the h

Question

An object of mass 500 g is projected in the air at an angle of 30 deg with the horizontal direction with a velocity of 10 m/s.

(a) The weight of the object; w = mg, m has to be in kg = g/1000

(b) the x –component (horizontal component) of the initial velocity: vxi = vi cos

(c) the y –component (vertical component) of the initial velocity: vyi = vi sin y vi = 10 m/s = 30 deg x

(d) the height the object rises: vfy2 = viy 2 + 2gy (e) time to reach the highest height: vfy = viy - g t (f) time of flight: Twice the time to reach the highest height

(g) the range of the projectile: vi 2 sin 2i/g or (vxi) x (time of flight)

(h) the velocity at the highest point, vyf

(i) the velocity just before touching the ground, vi (j) the kinetic energy at the moment of projection: ½ mvi 2

(k) the kinetic energy at the highest height, ½ mvxi 2

(l) the potential energy at the point of projection: mgh = ?

(m) the potential energy at the highest height: mgh = ?

(n) the angle of projection for which the range will the maximum, 2=90o

(o) the maximum range vi 2 /g=?

Explanation / Answer

As we know that

(a)

The weight of the object = w = mg = 0.500*9.8 = 4.9 N

(b)

the x –component (horizontal component) of the initial velocity: vxi = vi cos

the x –component (horizontal component) of the initial velocity: vxi = 10*cos30 = 8.660254 m/s

(c)

the y –component (vertical component) of the initial velocity: vyi = vi sin

the y –component (vertical component) of the initial velocity: vyi = 10*sin30

the y –component (vertical component) of the initial velocity: vyi = 5 m/s

(d)

the height the object rises: vfy2 = viy 2 + 2gy

As we know that at highest point

vfy = 0 m/s

the height the object rises: 0 = 5^2 - 2*9.8*y

y = the height the object rises = 1.275510204 m

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