Question

a body is thrown with an initial velocity of 10 m/s at an angle of 37 degree horizontally. find the position of vector after 1 sec

please explain the answer...

correct = brainleist
wrong= report​

Answer 1

(i) 1.2 sec, (ii) 1.8 m, (iii) 9.6 m, (iv) (16iˆ−8jˆ)−(8iˆ+jˆ)

Solution

(i) time of flight T=2usinθg=2×10×3510=65=1.2 sec

(ii) Maximum height H=u2sin2θ2g=100(925)2×10=95=1.8m

(iii) R=u2sin2θg=100×2×35×4510=24025=9.6m

(iv) x=8×1=8m

y=6×1−12×10×(1)2=1m

r→=8iˆ+jˆ

Caution : This equation does ot given the horizontal distance travelled by a projectile when the final height is not the launch height.

Maximum Range

R=u2sin2θg

for θ=45∘, R is maximum

Rmax=u2g