Two launch angles for an approaching target

Cannonball trajectorySource-faithful qualitative diagram; distances are schematic unless an exact function is stated.xyOCannonball (x,y)θ°Ground
Question diagram 1
Cannonball trajectorySource-faithful qualitative diagram; distances are schematic unless an exact function is stated.xyOCannonball (x,y)θ°Ground

Given cannonball trajectory, launch angle, axes, point(x,y) and ground.

The diagram shows the trajectory of a cannonball fired off from an origin OO with an initial speed of v m s−1v\,\mathrm{m\,s^{-1}} and at an angle of θ∘\theta^\circ above the ground. At time tt seconds, its position can be modelled by x=(vcos⁡θ)tx=(v\cos\theta)t, y=(vsin⁡θ)t−5t2y=(v\sin\theta)t-5t^2, where xx m is horizontal distance with respect to OO and yy m is vertical distance with respect to ground level.

(i) Find the horizontal distance, dd m, that a cannonball would have travelled by the time it hits the ground. Leave your answer in terms of vv and θ\theta. marks

Use v=200v=200 to answer the remaining parts. An approaching target is travelling at a constant speed of 10 m s−110\,\mathrm{m\,s^{-1}} along the ground. A cannonball is fired towards it when it is 3000 m away. You may assume its height is negligible.

(ii) Show that in order to hit the target, the possible angles at which the cannonball should be fired are 22.7∘22.7^\circ and 69.5∘69.5^\circ. marks

(iii) Explain at which angle the cannonball should be fired in order to hit the target earlier. marks

(iv) Given θ=22.7\theta=22.7, find the angle that the tangent to the trajectory makes with the horizontal when x=370x=370. marks

Not printed in supplied question

Total marks: 12

Solution

(i)

Projectile range

x=(vcos⁡θ)tx=(v\cos\theta)t, y=(vsin⁡θ)t−5t2y=(v\sin\theta)t-5t^2. Find the horizontal distance dd travelled by the time the cannonball hits the ground, in terms of vv and θ\theta.

Hint

  1. The ball hits the ground when y=0y=0 with t>0t>0.

  2. Substitute that tt into xx.

Answer · (i)

d=v2sin⁡θcos⁡θ/5d=v^2\sin\theta\cos\theta/5 metres.

(ii)

Moving target

v=200v=200. A target moves along the ground at 10 m s−110\ \mathrm{m\,s^{-1}} towards the gun and is 3000 m away when the ball is fired. Show that the possible angles are 22.7∘22.7^\circ and 69.5∘69.5^\circ.

Explore this idea

Hint

  1. The ball lands after t=vsin⁡θ5t=\frac{v\sin\theta}5 seconds at distance dd.

  2. In that time the target covers 10t10t, so it must be at 3000−10t=d3000-10t=d.

Answer · (ii)

θ=22.7∘\theta=22.7^\circ or 69.5∘69.5^\circ.

Explore the idea

Race the two shots against the target

Launch both possibilities on the same clock. Pause at either landing to see why the lower shot hits sooner.

Two projectile trajectories and a moving ground targetCircle: low shot. Diamond: high shot. Square: target. Positions, velocities and landing times are given below the controls.00504750100915001513225020183000Height (m)Distance from launch (m)
Shared time0.00 s
Playback speed
Low shot · 22.7°
0.0 m above ground

x = 0.0 m · vertical velocity = 77.1 m/s · tangent angle = 22.7°

High shot · 69.5°
0.0 m above ground

x = 0.0 m · vertical velocity = 187.3 m/s · tangent angle = 69.5°

Target position: 3000.0 m. Both shots are still in flight.

Angles are displayed rounded; calculations use the unrounded solutions. Each shot is a separate scenario. Landed markers stay at their impact points while the clock continues for the other shot. Axes have different scales; read the numerical tangent angle below.

(iii)

Comparing times

Explain at which angle the cannonball should be fired in order to hit the target earlier.

Hint

  1. The time of flight is t=vsin⁡θ5t=\frac{v\sin\theta}5.

  2. Compare sin⁡22.7∘\sin22.7^\circ and sin⁡69.5∘\sin69.5^\circ.

Answer · (iii)

22.7∘22.7^\circ hits earlier: about 15.42 s versus 37.46 s.

(iv)

Parametric tangent

Given θ=22.7∘\theta=22.7^\circ, find the angle that the tangent to the trajectory makes with the horizontal when x=370x=370.

Explore this idea

Hint

  1. Find the time when x=370x=370.

  2. dydx=dy/dtdx/dt\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.

  3. The angle is tan⁡−1\tan^{-1} of the gradient.

Answer · (iv)

17.2∘17.2^\circ above horizontal.

Use the specified rounded launch angle 22.7° and degree mode.

Explore the idea

The tangent as the ball flies

Slide the point along the 22.7° trajectory. Jump to x = 370 to read the gradient.

Position along the curve(370, 134.66747)t = 2.00534
Try an example
Tangent gradient
0.30962

At x=370x=370 the gradient is 0.30960.3096: tan⁡−10.3096=17.2∘\tan^{-1}0.3096=17.2^\circ above the horizontal.

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • Trajectory
  • Tangent
Key idea

The tangent angle falls steadily from the launch angle 22.7∘22.7^\circ to 00 at the peak, then goes negative as the ball descends.

Ideas to try
  1. At the peak, what is the gradient?
  2. Slide past the peak: the gradient changes sign.