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A projectile is thrown with an initial v...

A projectile is thrown with an initial velocity of `vecv = (phati+qhatj)` m/s. If the range of the projectile is four times the maximum height reached by it, then

A

p = 2q

B

q = 4p

C

q - 2p

D

q = p

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The correct Answer is:
To solve the problem, we need to establish the relationship between the range and the maximum height of a projectile thrown with an initial velocity given in vector form. Let's break it down step by step. ### Step 1: Identify the Components of Initial Velocity The initial velocity of the projectile is given as: \[ \vec{v} = p \hat{i} + q \hat{j} \text{ m/s} \] Here, \( p \) is the horizontal component (velocity in the x-direction), and \( q \) is the vertical component (velocity in the y-direction). ### Step 2: Formula for Maximum Height The maximum height \( H \) reached by a projectile is given by the formula: \[ H = \frac{v_y^2}{2g} \] Substituting \( v_y = q \): \[ H = \frac{q^2}{2g} \] ### Step 3: Formula for Range The range \( R \) of the projectile is given by the formula: \[ R = \frac{2v_x v_y}{g} \] Substituting \( v_x = p \) and \( v_y = q \): \[ R = \frac{2pq}{g} \] ### Step 4: Relationship Between Range and Maximum Height According to the problem, the range is four times the maximum height: \[ R = 4H \] Substituting the expressions for \( R \) and \( H \): \[ \frac{2pq}{g} = 4 \left(\frac{q^2}{2g}\right) \] ### Step 5: Simplifying the Equation We can simplify the equation: \[ \frac{2pq}{g} = \frac{4q^2}{2g} \] This simplifies to: \[ \frac{2pq}{g} = \frac{2q^2}{g} \] Multiplying both sides by \( g \) (assuming \( g \neq 0 \)): \[ 2pq = 2q^2 \] ### Step 6: Dividing by 2 Dividing both sides by 2: \[ pq = q^2 \] ### Step 7: Rearranging the Equation Rearranging gives: \[ pq - q^2 = 0 \] Factoring out \( q \): \[ q(p - q) = 0 \] ### Step 8: Solving the Equation This gives us two possible solutions: 1. \( q = 0 \) (which is not a valid solution for projectile motion) 2. \( p - q = 0 \) or \( p = q \) ### Conclusion Thus, the relationship between the components of the initial velocity is: \[ p = q \]

To solve the problem, we need to establish the relationship between the range and the maximum height of a projectile thrown with an initial velocity given in vector form. Let's break it down step by step. ### Step 1: Identify the Components of Initial Velocity The initial velocity of the projectile is given as: \[ \vec{v} = p \hat{i} + q \hat{j} \text{ m/s} \] Here, \( p \) is the horizontal component (velocity in the x-direction), and \( q \) is the vertical component (velocity in the y-direction). ...
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