Question

A stone is thrown in air with velocity of 20 m/sec at an angle of 30o to the horizontal. Find its ( g = 10m/ 2 sec )
1) Maximum height
2) Time of flight
3) Range
4) Maximum range

Answer 1

Answer:

Hiee!!

Explanation:

I am going to assume this is a high school homework question, and that we are therefore assuming no air resistance or aerodynamic effects of any sort. Probably this is not the best place to ask this question, but I will answer anyway.

You can treat the vertical and horizontal components of velocity separately, then find the velocity using Pythagoras’ theorem.

The horizontal component (lets call it vh) will just be a constant 20 m/s

The vertical component (lets call it vv) can be given as a function of time, treating acceleration as a constant we have

vv(t|a) = vvi + at

where vv is the vertical component of velocity, vvi is the initial vertical velocity, t is time, and a is acceleration.

vi is zero as the stone is thrown horizontally (has no vertical component to its velocity). If we assume that we are in earths gravity, then a=g =-9.8 ms^-2. t=3 by the question specification.

The total velocity will be given by the combination of the horizontal and vertical components, and can be easily found using pythagoras’s theorem.

v=(vh^2 + vv^2)^0.5

Substituting we have

v=[20^2 + (3*-9.8)^2]^0.5 =35.56 ms^-1

If you want to give velocity as a polar coordinate, then we also need to find the angular component to the velocity vector, given in degrees by (I really need a diagram to explain this but maybe you can work it out on paper) the following -if we take the initial direction of the stones path as zero degrees, and move anti-clockwise, we then have:

v(theta) = tan(vv/vh)= 272.7 degrees

then in polar coordinate form we have

V=[v;v(theta)] = [35.56 ms^-1; 272.7 degrees]

If you additionally want the position of the stone, then we can use

sh=shi + vhi.t + 1/2 aht^2

sv=svi + vhi.t + 1/2 avt^2

where shi is the original horizontal position, vhi is the initial horizontal position, aht is the acceleration in the horizontal dimension, etc.

let shi=0; svi=0 (i.e. the thrower is our point of origin)

Then

sh=0 + 20*3 + 0 = 60m

sv=0 + 0 + 1/2*-9.8*3^2 = -44.1m

Then we can fully define the stone’s properties by the four dimensional vector P:

P=[v;v(theta);sh;sv]= [35.56 ms^-1; 272.7 degrees; 60m, -44.1m]