Question

A hi-tech high-powered catapult is located at point A and is aimed at an angle of = 45 ° with the horizontal.

The projectile launched from the catapult is just able to clear the peak of the mountain at the top of its

trajectory. The elevation of the catapult is 423 m above sea level and the distance d is 4031 m.

Find the following:

(a) The velocity V₀ that the projectile leaves the catapult (in m/s),

(b) The time it takes the projectile to reach the peak of its trajectory (in seconds),

(c) The maximum height above sea level, h, the projectile reaches (in metres),

(d) The time it takes for the projectile to hit the water (in seconds), and

(e) The range, R, of the projectile (in metres).

Answer 1

a)

Given,

Elevation of catapult above sea level, H = 423 m

distance between catapault and mountain peak, d = 4031 m

Since max height is reached at the peak of mountain hence distance between catapult and mountain is half the range

hence, Range, R = 2* 4031 = 8062 m

Range, R = V₀² sin2 / g = V₀² sin90 / 9.8

=> 8062 = V₀² / 9.8

=> V₀ = 281.08 m/s

b)

Time taken to reach the peak is half the time of flight

t = V₀ sin / g

=> t = 281.08 * sin45 / 9.8 = 281.08 * 0.707 / 9.8 = 20.278 s

c)

Now

max height attained by the trajectory, H = V₀² sin² / 2g

=> H = (281.08)² * (0.707)² / 2*9.8

=> H = 2015.45 m

Total height gained above sea level = H + height above sea level = 2015.45 + 423

                                                           = 2438.45 m

d)

Time taken to hit water = 2*time taken to reach max height + time taken to reach water from point B

let time taken from point B to reach water be t

h₁ = 423 m

As we know initial speed of trajectory is equal to finl speed of trajectory at point B

hence speed along y axis , v = V₀ sin = 281.08 * sin45 = 281.08 * 0.707 = 198.72 m/s

Now

h = vt+ 1/2*g*t²

=> 423 = 198.72*t +0.5*9.8*t²

=> 4.9*t² + 198.72*t - 423 = 0

Solving this we get

t = 2.03 s

Hence total time taken to reach water = 2*20.278 + 2.03

                                                               = 42.583 s

e)

Horizontal distance travelled after point B = V₀ cos * t = 281.08 * cos45 * 2.03

                                                                      = 281.08 * 0.707 * 2.03

                                                                      = 403.41 m

Total horizontal distance travelled = 8062 + 403.41 = 8465.41 m