143761 The speed of a projectile at its maximum height is \(\frac{\sqrt{3}}{2}\) times its initial speed. If the range of the projectile is \(P\) times the maximum height attained by it, then \(P\) equals

143762 An object is projected with a velocity of \(20 \mathrm{~ms}^{-1}\) making an angle of \(45^{\circ}\) with horizontal. The equation for the trajectory is \(h=\mathbf{A x}-\mathbf{B x}^{2}\), where \(h\) is height, \(x\) is horizontal distance \(A\) and \(B\) are constants. The ratio \(A: B\) is \((g=10\) \(\mathbf{m s}^{-2}\) )

143763 The horizontal and vertical displacements of a projectile at time \(t\) are \(x=36 t\) and \(y=48 t-\) \(4.9 \mathrm{t}^{2}\) respectively. Initial velocity of the projectile in \(\mathrm{ms}^{-1}\) is

143764 The horizontal and vertical displacements \(x\) and \(y\) of a projectile at a given time \(t\) are given by \(x=6 t\) metre and \(y=8 t-5 t^{2}\) metre. The range of the projectile in metre is

143765 The equation of trajectory of a projectile is
\(y=10 x-\left(\frac{5}{9}\right) x^{2}\)
If we assume \(\mathrm{g}=10 \mathrm{~ms}^{-2}\), the range of projectile (in metre) is

143761 The speed of a projectile at its maximum height is \(\frac{\sqrt{3}}{2}\) times its initial speed. If the range of the projectile is \(P\) times the maximum height attained by it, then \(P\) equals

143762 An object is projected with a velocity of \(20 \mathrm{~ms}^{-1}\) making an angle of \(45^{\circ}\) with horizontal. The equation for the trajectory is \(h=\mathbf{A x}-\mathbf{B x}^{2}\), where \(h\) is height, \(x\) is horizontal distance \(A\) and \(B\) are constants. The ratio \(A: B\) is \((g=10\) \(\mathbf{m s}^{-2}\) )

143763 The horizontal and vertical displacements of a projectile at time \(t\) are \(x=36 t\) and \(y=48 t-\) \(4.9 \mathrm{t}^{2}\) respectively. Initial velocity of the projectile in \(\mathrm{ms}^{-1}\) is

143764 The horizontal and vertical displacements \(x\) and \(y\) of a projectile at a given time \(t\) are given by \(x=6 t\) metre and \(y=8 t-5 t^{2}\) metre. The range of the projectile in metre is

143765 The equation of trajectory of a projectile is
\(y=10 x-\left(\frac{5}{9}\right) x^{2}\)
If we assume \(\mathrm{g}=10 \mathrm{~ms}^{-2}\), the range of projectile (in metre) is

143761 The speed of a projectile at its maximum height is \(\frac{\sqrt{3}}{2}\) times its initial speed. If the range of the projectile is \(P\) times the maximum height attained by it, then \(P\) equals

143762 An object is projected with a velocity of \(20 \mathrm{~ms}^{-1}\) making an angle of \(45^{\circ}\) with horizontal. The equation for the trajectory is \(h=\mathbf{A x}-\mathbf{B x}^{2}\), where \(h\) is height, \(x\) is horizontal distance \(A\) and \(B\) are constants. The ratio \(A: B\) is \((g=10\) \(\mathbf{m s}^{-2}\) )

143763 The horizontal and vertical displacements of a projectile at time \(t\) are \(x=36 t\) and \(y=48 t-\) \(4.9 \mathrm{t}^{2}\) respectively. Initial velocity of the projectile in \(\mathrm{ms}^{-1}\) is

143764 The horizontal and vertical displacements \(x\) and \(y\) of a projectile at a given time \(t\) are given by \(x=6 t\) metre and \(y=8 t-5 t^{2}\) metre. The range of the projectile in metre is

143765 The equation of trajectory of a projectile is
\(y=10 x-\left(\frac{5}{9}\right) x^{2}\)
If we assume \(\mathrm{g}=10 \mathrm{~ms}^{-2}\), the range of projectile (in metre) is

143761 The speed of a projectile at its maximum height is \(\frac{\sqrt{3}}{2}\) times its initial speed. If the range of the projectile is \(P\) times the maximum height attained by it, then \(P\) equals

143762 An object is projected with a velocity of \(20 \mathrm{~ms}^{-1}\) making an angle of \(45^{\circ}\) with horizontal. The equation for the trajectory is \(h=\mathbf{A x}-\mathbf{B x}^{2}\), where \(h\) is height, \(x\) is horizontal distance \(A\) and \(B\) are constants. The ratio \(A: B\) is \((g=10\) \(\mathbf{m s}^{-2}\) )

143763 The horizontal and vertical displacements of a projectile at time \(t\) are \(x=36 t\) and \(y=48 t-\) \(4.9 \mathrm{t}^{2}\) respectively. Initial velocity of the projectile in \(\mathrm{ms}^{-1}\) is

143764 The horizontal and vertical displacements \(x\) and \(y\) of a projectile at a given time \(t\) are given by \(x=6 t\) metre and \(y=8 t-5 t^{2}\) metre. The range of the projectile in metre is

143765 The equation of trajectory of a projectile is
\(y=10 x-\left(\frac{5}{9}\right) x^{2}\)
If we assume \(\mathrm{g}=10 \mathrm{~ms}^{-2}\), the range of projectile (in metre) is

143761 The speed of a projectile at its maximum height is \(\frac{\sqrt{3}}{2}\) times its initial speed. If the range of the projectile is \(P\) times the maximum height attained by it, then \(P\) equals

143762 An object is projected with a velocity of \(20 \mathrm{~ms}^{-1}\) making an angle of \(45^{\circ}\) with horizontal. The equation for the trajectory is \(h=\mathbf{A x}-\mathbf{B x}^{2}\), where \(h\) is height, \(x\) is horizontal distance \(A\) and \(B\) are constants. The ratio \(A: B\) is \((g=10\) \(\mathbf{m s}^{-2}\) )

143763 The horizontal and vertical displacements of a projectile at time \(t\) are \(x=36 t\) and \(y=48 t-\) \(4.9 \mathrm{t}^{2}\) respectively. Initial velocity of the projectile in \(\mathrm{ms}^{-1}\) is

143764 The horizontal and vertical displacements \(x\) and \(y\) of a projectile at a given time \(t\) are given by \(x=6 t\) metre and \(y=8 t-5 t^{2}\) metre. The range of the projectile in metre is

143765 The equation of trajectory of a projectile is
\(y=10 x-\left(\frac{5}{9}\right) x^{2}\)
If we assume \(\mathrm{g}=10 \mathrm{~ms}^{-2}\), the range of projectile (in metre) is