Question

Can someone explain with steps please​

Answer 1

Here

initial velocity=u

• u1=2.5m/s
• u2=?
• $\theta_1=65°$
• $\theta_2=15°$

Now

$\\ \sf\longmapsto R_1=R_2$

$\\ \sf\longmapsto \dfrac{u_1^2sin2\theta_1}{g}=\dfrac{u_2^2sin2\theta_2}{g}$

$\\ \sf\longmapsto \dfrac{(2.5)^2.sin2(65)}{g}=\dfrac{u_2^2sin2(15)}{g}$

• Cancel g

$\\ \sf\longmapsto 6.25sin130=u_2^2sin30$

$\\ \sf\longmapsto 6.25(0.76)=u_2^2(0.5)$

$\\ \sf\longmapsto 5.13=0.5u_2^2$

$\\ \sf\longmapsto u_2^2\approx 10$

$\\ \sf\longmapsto u_2\approx 3.1m/s$

Option B is correct

Answer 2

Answer:

The speed of other projectile is $3.1m/s$

Explanation:

Range of projectile is given by the equation

$\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}$

Here we have same range

Hence

$\frac{\mathrm{2.5}^{2} \cdot \sin (2 \times 65)}{\mathrm{g}}=\frac{\mathrm{u}^{2} \cdot \sin (2 \times 15)}{\mathrm{g}}\\\\u^2=\frac{2.5^2\sin130}{\sin30} \\\\u=3.10m/s$