Question

A stone is projected in space at an angle of 45 to horizontal at an initial velocity of 10m/s. find the range of projectile

Answer 1

The final resultant equation is:

y=s0y+(x−s0x)tanθ−g2u2cos2θ(x−s0x)2

Where y is pointing up and x is to the right. Theta is the initial launch angle, g is gravitation and u is the initial velocity.

If we place the starting position of the ball at:

(s0x,s0y)=(0,0)

And we observe that:

θ=45∘

Then the equation of the trajectory is:

y=x−9.81u2x2

So the ball therefore is at ground level at launch (x=0) and once it hits the ground again:

0=(1−9.81u2x)x

x=0,u29.81

The wall is 6m prior to the touchdown point of the ball (in other words x=u29.81−6) and the ball ‘just clears’ it. Therefore, we can assume the top of the wall coincides with the trajectory. And therefore, the y value at this position is the height of the wall:

y=(1−9.81u2(u29.81−6))(u29.81−6)

y=6−369.81u2

From this equation, it's clear that solving this problem is dependent on the initial launch speed. Which is unknown. We also know that the minimum speed this situation is possible in is where y>0 which puts a lower bound on initial speed.

And it makes sense thar we can't determine this height without the initial speed. If the speed was too low but above the minimum, the wall could be right near the launch location and still be 6m from the touchdown point, therefore being a very short height. Similarly, if the initial speed was large, the height could be much larger.

Answer 2

$\huge {\bold {\underline {\underline{Given : - }}}}$

• Intial velocity amd angle made by the stone with the horizontal are 10m/s and 45°

$\huge {\bold {\underline{ \underline{To \: find : - }}}}$

• Range of the projectile

$\huge{ \bold{ \underline{ \underline{ Solution : - }}}}$

We have ,

• u = 10m/s
• θ = 45°
• g = 9.8 m/s²

Range of the projectile is given by ,

$\dag \: \large {\underline {\bold {\boxed{R = \frac{ {u}^{2} \sin(2 \theta) }{g} }}}}$

$: \implies \: R \: = \frac{ {(10)}^{2} \ \times \sin(2 \times 45) }{9.8} \\ \\ : \implies \: R \: = \frac{100 \times \sin(90) }{9.8} \\ \\ \large { \bold {\boxed { \sin(90 {}^{0} ) = 1}}} \\ \\ : \implies R = \frac{100 \times 1}{9.8} \\ \\ : \implies \: R \: = \frac{100}{9.8} \\ \\ : \implies \: R = 10.2 \: m$

$\dag \: \rm \: Hence ,\: the \: range \: of \: the \: given \: projectile \: is \: 10.2 \: m$

$\huge{ \bold {\underline {\underline{ additional \: info : - }}}}$

★ Maximum height reached by the projectile is given by ,

$\large { \underline{\bold {\boxed{h = \frac{ {u}^{2} \sin {}^{2} ( \theta) }{2g} }}}}$

Where ,

• u is initial velocity
• g is acceleration due to gravity
• θ is angle made with the horizontal

★ Time of flight of the projectile is given by ,

$\large{\underline{\bold{\boxed{{T}_{f}\:=\:\frac{2usin(\theta)}{g}}}}}$

where ,

• u is initial velocity
• g is acceleration due to gravity
• θ is angle made with the horizontal