Solveeit Logo

Question

Physics Question on projectile motion

A projectile is thrown from a point OO on the ground at an angle 4545^{\circ} from the vertical and with a speed 52m/s5 \sqrt{2} \,m / s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 05s05\, s after the splitting. The other part, tt seconds after the splitting, falls to the ground at a distance xx meters from the point OO. The acceleration due to gravity g=10m/s2g =10 \,m / s ^{2}.
The value of xx is ______

Answer

X=3R2X = \frac{3R}{2} as Xcm = R

R=u2sin(2θ)gR = \frac{{u_2 \sin(2\theta)}}{g}

=5010=5= \frac{50}{10} = 5

X=3R2=152=7.5m⇒X = \frac{3R}{2} = \frac{15}{2} = 7.5 m
Answer: 7.50