A particle is projected horizontally with velocity V m/s, from a point h meters above ground. Taking g ms^-2 as the acceleration due to gravity,
1. Show the equation of the path is given by y = (2hV^2 - gx^2)/ 2V^2
2. Find the range of the particle.
Thanks
To show the equation of the path, we can use the equations of motion for vertical and horizontal motion separately.
1. Vertical motion:
Since the particle is projected horizontally, its initial vertical velocity is zero. The only force acting on the particle is gravity, so the equation of motion for vertical motion can be represented as:
y = h - (1/2)gt^2
However, we can express time (t) in terms of horizontal motion using the equation:
t = x/V (since the initial horizontal velocity is V)
Substituting the value of t in the vertical motion equation, we get:
y = h - (1/2)g(x/V)^2
2. Horizontal motion:
Since the particle is projected horizontally, there is no horizontal acceleration. Therefore, the horizontal velocity remains constant throughout the motion. The equation of motion for horizontal motion is:
x = Vt
Now, substitute the value of t from the equation t = x/V into the equation y = h - (1/2)g(x/V)^2:
y = h - (1/2)g[(x/V) * (x/V)]
Simplifying further, we get:
y = (2hV^2 - gx^2)/ 2V^2
This is the equation of the path for the particle.
To find the range of the particle, we need to determine the maximum horizontal distance it travels before hitting the ground. The range (R) can be calculated using the equation:
R = V * t
Since the particle is projected horizontally, the time taken (t) to reach the ground is the same as the time for the particle to reach its maximum height (h). We can calculate this time using the vertical equation of motion:
h = (1/2)g * t^2
Rearranging the equation, we get:
t = sqrt(2h/g)
Substituting the value of t into the equation for the range:
R = V * sqrt(2h/g)
Therefore, the range of the particle is given by R = V * sqrt(2h/g).