The figure shows a hill that is inclined at an angle ? = 45°...

The figure shows a hill that is inclined at an angle ? = 45° relative to the vertical. A projectile is launched from the origin 0 with an initial velocity v? of norm v? = 4 m/s. The velocity vector makes an angle of ? = 30° with the z-axis. With this information, determine the following: (a.) Find the maximum separation between the projectile and the hill. (b.) Find the distance between the origin and the point of landing of the projectile on the hill. Note: Use g = 9.8 m/s².

Transcript

00:01 Hi students, the components of acceleration on the particle, ax equal to g goes alpha and a is equal to minus g sine alpha.

00:17 Initial components of velocity, v0x is equal to v0 sine theta and v0 z equal to v0 cos theta.

00:32 Using the kinematic equations of motion, the x and z components of velocity as a function of time, vx is equal to v0x plus a xt implies vx equal to v0 sine theta plus g cos alpha t and vz equal to v0.

01:04 Z plus a zz of t which implies vzz equal to v0 cos theta minus g sine alpha t let this be equation number one and this be two and the displacement as a function of time x is equal to v0 x plus a x t square by two x x is equal to v0 x plus a x t square by two x x is equal to to v0 sine theta into t plus g cosa alpha t square by 2 that is this equation number 3 and is that equal to v0 cos theta into t plus minus of g sine alpha t square by 2 that can be written as v0 cost theta into t minus g sine alpha t square by 2 this be equation number 4 again components of velocity as a function of distance is v x square equal to v0 x square plus 2 a x into x therefore vx square equal to v x x squared equal to v0 sine theta the whole square plus 2g cos alpha x this is equation number 5 and vz square equal to v0 cos theta the whole square minus 2 g sine alpha z this is equation number 6 for the a part at maximum separation between the projectile and the hill, z will be z max and vz equal to 0.

03:26 From equation number 6, 0 equal to v0 cos theta the whole square minus 2g sine alpha z max where z max is equal to v0 cost theta of the whole square divided by 2g sine alpha which is equal to 4 into cos 30 the whole square divided by 2 into 9 .8 into sine 45 which on solving is obtained as 0 .87 meters.

04:13 Moving on to the b part, let the projectile lands at time t equal to t...

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