Question

Difficulty: MediumProjectile Motion

A cannonball is fired from ground level with an initial speed of 50 m/s50\text{ m/s} at an angle θ\theta above the horizontal such that tanθ=43\tan \theta = \frac{4}{3}. Assuming acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2 and neglecting air resistance, what is the magnitude of the velocity of the cannonball at the highest point of its trajectory?

  1. 30 m/s30\text{ m/s}Answer
  2. B
    0 m/s0\text{ m/s}
  3. C
    40 m/s40\text{ m/s}
  4. D
    50 m/s50\text{ m/s}

Answer

30 m/s30\text{ m/s}
At the maximum height of a projectile's flight, the vertical velocity component drops to zero due to gravity, while the horizontal component remains unchanged because horizontal acceleration is zero. For a launch speed of 50 m/s50\text{ m/s} at an angle with tanθ=43\tan \theta = \frac{4}{3}, cosθ=35\cos \theta = \frac{3}{5}. The magnitude of the velocity at apex is therefore equal to ux=50×35=30 m/su_x = 50 \times \frac{3}{5} = 30\text{ m/s}.

Step-by-Step Solution

1
Resolve the initial velocity vector into orthogonal horizontal and vertical components.
Given tanθ=43\tan \theta = \frac{4}{3}, the trigonometric ratios are cosθ=35\cos \theta = \frac{3}{5} and sinθ=45\sin \theta = \frac{4}{5}. Thus, ux=50×35=30 m/su_x = 50 \times \frac{3}{5} = 30\text{ m/s} and uy=50×45=40 m/su_y = 50 \times \frac{4}{5} = 40\text{ m/s}.
Resolving 2D projectile motion into independent orthogonal components simplifies kinematic analysis.
2
Determine the velocity components at the apex (highest point) of the trajectory.
At maximum height, the vertical velocity component vy=0 m/sv_y = 0\text{ m/s}. Since there is no horizontal acceleration, the horizontal velocity component remains vx=ux=30 m/sv_x = u_x = 30\text{ m/s}.
Gravity acts purely vertically, reducing vertical velocity to zero at the apex while leaving horizontal velocity unchanged.
3
Calculate the magnitude of total velocity at maximum height.
v=vx2+vy2=302+02=30 m/sv = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}.
The magnitude of a 2D velocity vector is determined by combining its orthogonal components using the Pythagorean formula.

Key Concept

Velocity components at maximum height in projectile motion
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