Question

The invention of the cannon in the fourteenth century made the catapult unnecessary and ended the safety of castle walls. Stone walls were no match for balls shot from cannons. Suppose a cannonball of mass 5.00 kg is launched from a height of 1.10 m, at an angle of elevation of 30.0° with an initial velocity of 51.3 m/s, toward a castle wall of height 30.0 m and located 219 m away from the cannon. The range of a projectile is defined as the horizontal distance traveled when the projectile returns to its original height. What will be the range reached by the projectile if it is not intercepted by the wall? (b) If the cannonball travels far enough to hit the wall, find the height at which it strikes.

Answer 1

Answer:

Explanation:

Given

mass of cannon=5 kg

Initial launch height =1.10 m

launch angle $\theta =30$

initial velocity=51.3 m/s

height of wall=30 m

distance of wall=219 m

Range of Projectile $=\frac{u^2\sin 2\theta }{g}$

$R=\frac{51.3^2\sin 60}{9.8}=232.56 m$

Trajectory of Projectile is given by

$y=x\tan \theta -\frac{gx^2}{2u^2\cos^2\theta }$

For x=219 m

$y=219\tan (30)-\frac{9.8\times 219^2}{2\times 51.3^2\times (\cos 30)^2}$

$y=\frac{219}{\sqrt{3}}-119.066$

$y=126.439-119.066=7.37$

i.e. cannon will hit at height of 7.37 m w.r.t initial launch

Height from ground 7.37+1.1=8.47 m

Thus cannon will strike at a height of 8.47 m from ground

Answer 2

The cannon ball will travel a range of 232.56 m and will strike the wall at a height of 8.47 m from ground.

What is the range of a projectile?

The range of a projectile is defined as the horizontal distance traveled when the projectile returns to its original height.

From the data provided:

• mass of cannon=5 kg
• Initial launch height =1.10 m
• angle from the horizontal = 30°
• initial velocity=51.3 m/s
• height of wall = 30 m
• distance of wall=219 m

The range of a projectile is calculated using the formula below:

$Range of Projectile =\frac{u^2\sin 2\theta }{g}$

$

R=\frac{51.3^2\sin 60}{9.8}=232.56 m$

Therefore, the range of the projectile is 232.56 m.

The trajectory of Projectile is given by:

$y=x\tan \theta -\frac{gx^2}{2u^2\cos^2\theta }$

For x=219 m

$y=219\tan (30)-\frac{9.8\times 219^2}{2\times 51.3^2\times (\cos 30)^2}$

$y=7.37$

Thus, the cannon will hit at height of 7.37 m based on initial launch

Height from ground 7.37+1.1 = 8.47 m

Therefore, the cannonball will strike the wall at a height of 8.47 m from ground.

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