Kinematics

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224 of 308

Question 224 of 308

The range of the projectile projected at an angle of 15∘{15}^{^{\circ}} with horizontal is 50m50m. If the projectile is projected with same velocity at an angle of 45∘{45}^{^{\circ}} with horizontal, then its range will be

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Answer: (A) 100m100m

The data given is
θ1=15∘θ2=45∘R1=50m{\theta }_{1}=15^{\circ} {\theta }_{2}=45^{\circ} {R}_{1}=50m
The formula for range of a projectile is given by
R=u2sin⁡2θg...(i)R=\dfrac{{u}^{2}\sin 2\theta }{g}...(i)
Substituting the values in equation (i)
R1=u2sin⁡2θ1g⇒50=u2sin⁡30∘g⇒u2g=100{R}_{1}=\dfrac{{u}^{2}\sin 2{\theta }_{1}}{g} \Rightarrow 50=\dfrac{{u}^{2}\sin 30^{\circ}}{g} \Rightarrow \dfrac{{u}^{2}}{g}=100
Let the new range be R2{R}_{2}. The magnitude of the range is
R2=u2sin⁡2θ2g⇒R2=(u2g)sin⁡90∘⇒R2=100m{R}_{2}=\dfrac{{u}^{2}\sin 2{\theta }_{2}}{g} \Rightarrow {R}_{2}=\left(\dfrac{{u}^{2}}{g}\right)\sin 90^{\circ} \Rightarrow {R}_{2}=100m

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