Motion in a Plane: A Projectile's Trajectory?
A particle is projected at an angle of 60° to the horizontal with an initial velocity of 20 m/s. If the particle lands 50 m away from the point of projection, what is the time it took to reach the ground, assuming a constant acceleration due to gravity of 9.8 m/s²?
1 Answer
📌 CONCEPT: The trajectory of a projectile is a parabola that can be described using the equations of motion under constant acceleration due to gravity.
📐 RULE / FORMULA: For a projectile launched at an angle θ with an initial velocity u, the horizontal range R can be calculated using the formula R = (u² * sin(2θ)) / g, where g is the acceleration due to gravity.
💡 WORKED EXAMPLE: Given u = 20 m/s, θ = 60°, and R = 50 m, we can use the formula to find the time taken to reach the ground. First, we calculate the vertical component of the initial velocity, v0y = u * sin(θ) = 20 * sin(60°) = 17.32 m/s. Then, using the equation for time of flight t = (2 * v0y) / g, we get t = (2 * 17.32) / 9.8 = 3.54 s.
⚠️ COMMON MISTAKE: Students often forget to consider the vertical component of the initial velocity when calculating the time of flight, and instead use the horizontal component or the entire initial velocity, leading to incorrect results.
05 Sept 26
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