A cannon ball is launched 25 m/s 53 degrees above the horizontal to the right. Find the maximum height it can reach, horizontal distance it covers and total time of flight.

Answer 1

Remember below three equations that can be used in 1D and 2D kinematics. Up/right direction is positive and down/left direction is negative. We need to use these equations separately in vertical and horizontal directions.

d = distance traveled or Δx or Δy, v = final velocity or v_f, u = initial velocity or v_{0 ,}

a = acceleration or g, Δt = time difference

d = uΔt + 0.5aΔt²

v² = u² + 2ad

v = u + aΔt

a)

Here initial velocity is 25 m/s at 53⁰

a = g = -9.8 m/s (g is acceleration due to gravity. It is pointing downwards, so it is negative).

Cannon ball has initial vertical velocity of u_y = 25 sin(53⁰) = 19.966 m/s

Cannon ball has a final vertical velocity of v_y = 0 m/s at maximum height h

v² = u² + 2ad ...apply this equation to vertical direction where d = height h

0 = 19.966² + 2 (-9.8) h

0 = 398.64 - 19.6h ...add 19.6h to both sides

19.6h = 398.64 ...divide both sides by 19.6

h = 20.4 m ... this is the max height reached by the cannon ball.

b)

Cannon ball has initial vertical velocity of u_y = 25 sin(53⁰) = 19.966 m/s

Cannon ball has final vertical velocity of v_y = -19.966 m/s when it reaches ground

v = u + aΔt ...apply this equation to vertical direction

-19.966 = 19.966 + (-9.8)Δt --- subtract 19.966 from both sides

-39.932 = -9.8Δt ...Divide both sides by -9.8

4.07 seconds = Δt, total time of flight

c)

Since there is no force acting in horizontal direction on cannon ball, its horizontal accelation = 0 m/s²

initial horizontal velocity u = 25 cos(53⁰⁾ = 15.05 m/s

Δt = time the ball travels horizontally = time the ball travels vertically (calculated above) = 4.07 seconds

d = uΔt + 0.5aΔt² ... d is the horizontal distance covered

d = 15.05 x 4.07 + 0.5 x 0 x 4.07²

^{d = 61.25 + 0}

^{horizontal distance it covers, d = 61.25 m}