Multiplicative inverse matrix generalized form to solve is
which of the following?
AA-1 = A-1A =
1
AA-2 = A-3A =
3AA4 = A-1A =
1
AA-1 = 2A-5A =
Q
Answers
Answer:
Answer:
Maximum height = 125 m
Time of flight = 5 seconds
Horizontal range = 250√3 m
Solution:
It is clear that the motion of the bullet can be categorised as a projectile motion. Let's find the required variables from the question,
Initial velocity, u = 100 m/s
Angle with the horizontal, = 30°
We have to find,
Maximum height reached by the bullet.
Time of flight.
Horizontal range.
For maximum height, we have the following formula,
⇒ Hₘₐₓ = u² sin²ϑ / 2g
⇒ Hₘₐₓ = (100)² (sin 30°)² / 2×10
⇒ Hₘₐₓ = 10⁴ (1/2)² / 20 [ sin 30° = 1/2 ]
⇒ Hₘₐₓ = 10³ / 2×4
⇒ Hₘₐₓ = 1000 / 8
⇒ Hₘₐₓ = 125 m
Now, Let's find Time of flight, which we is given by the formula,
⇒ T = (Initial velocity in vertical direction) / g
⇒ T = 2u sin ϑ / g
⇒ T = 2×100×sin 30° / 10 [ sin 30° = 1/2 ]
⇒ T = 20 × 1/2
⇒ T = 10 seconds
Furthermore, For getting the horizontal range, we use the following formula,
⇒ R = u² sin 2ϑ / g
⇒ R = (100)² × sin (2×30) / 10
⇒ R = 10000 × sin 60° / 10
⇒ R = 1000 × √3/2 [ sin 60° = √3/2 ]
⇒ R = 500√3 m