Question

A projectile is fired from ground level at time t=0, at an angle θ with respect to the horizontal. It has an initial speed v0 Assume the ground is level. Express answers in parts a - d in terms of v0 θ and g (the magnitude of the acceleration due to gravity.)
a) Find the time, tH, it takes the projectile to reach its maximum height H
b) Find tR, the time at which the projectile hits the ground after having traveled through a horizontal distance R
c) Find H, the maximum height attained by the projectile.
d) Find the total distance R(range) traveled in the x direction by the projectile.

Answer 1

Answer:

(a) Time to reach maximum height, $t_{H} =\frac{v_{0}\sin\theta }{g}$

(b) Time to reach the ground, $t_{R} =2t_{H} =\frac{2v_{0}\sin\theta }{g}$

(c) Maximum height, $H=\frac{(v_{0}\sin\theta) ^{2} }{2g}$

(d) Total horizontal distance, $R=\frac{v_{0} ^{2}\sin2\theta }{g}$

Explanation:

According to the problem, v₀ is initial speed, g is acceleration due to gravity and θ is angle with respect to the horizontal.

The horizontal and vertical components of the initial speed are v₀cosθ and v₀sinθ respectively.

There is no acceleration acting horizontally on the projectile but there is vertical acceleration g is acting on the projectile.

(a) The equation of motion along vertical direction is:

v = v₀sinθ - gt

At maximum height, the speed of projectile is zero i.e.v is zero in the above equation. Thus, the above equation becomes :

$0=v_{0}\sin\theta-gt_{H}$

$t_{H} =\frac{v_{0}\sin\theta }{g}$

(b) The time at which projectile hits the ground is :

$t_{R} =2t_{H} =\frac{2v_{0}\sin\theta }{g}$      ....(1)

(c) The equation of motion along vertical direction is :

v² = (v₀sinθ)² - 2gs

Here s is the vertical height of the projectile.

At maximum height H, v is zero. So, the above equation becomes :

$0=(v_{0}\sin\theta) ^{2}-2gH$

$H=\frac{(v_{0}\sin\theta) ^{2} }{2g}$

(d) The equation of motion along horizontal is :

x = (v₀cosθ)t

Here x is horizontal distance travel by projectile.

The total distance (R) traveled by projectile is :

$R=(v_{0}\cos\theta)t_{R}$

Substitute the equation (1) in above equation.

$R=\frac{2v_{0} ^{2} \sin\theta\cos\theta}{g}$

Using trigonometry relation:

2sinθcosθ = sin2θ

$R=\frac{v_{0} ^{2}\sin2\theta }{g}$