Question

Aarav hit a ball straight from a point a to point B it covered a distance of 30 M in first two minutes if internist and average velocity of the ball always remain the same the distance covered by the ball in the next 6 minutes would be​

Answer 1

Answer:

Here θ=30o,u=30ms−1

a. The time taken by the ball to reach the highest point is half the total time of flight. As the time of ascending and descending is same for a projectile without air resistance. the time to reach the highest point T

tH=2T=gusinθ=1030×sin30o=1.5s

b. The maximum height reached is

2gu2sin2θ=2g(302)×(sin30o)2=2×10×4900=11.25m

c. Horizontal range = gu2sin2θ

= 10(30)2sin2(30o)

= 209003=453m

d. The time for which thc ball is in air is same as its time of eight, i.e.,

g2usinθ=102×30×sin30o=3

Explanation:

mark me as brainliest pls