Physics projectile solve fast
Find the velocity of a projectile at the highest point, if it is projected with a speed 15 m s −1 , in the direction 45 o above horizontal. [take g=10ms −2 ]
Mordetor please solve
Answers
EXPLANATION.
The velocity of a projectile at the highest point.
It is projected with a speed = 15 m/s.
In the direction 45° above the horizontal.
⇒ g = 10 m/s².
As we know that,
In x - direction.
⇒ uₓ = ucosθ
⇒ aₓ = 0.
⇒ u = 15 m/s.
⇒ θ = 45°.
Put the values in the equation, we get.
⇒ uₓ = 15 x cos45°.
⇒ uₓ = 15 x 1/√2.
⇒ uₓ = 15/√2.
⇒ uₓ ≈ 10.606 m/s.
Velocity of particle at highest point = 10.606 m/s.
MORE INFORMATION.
Motion under gravity.
If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting air resistance then,
(1) Maximum height attained H = u²/2g.
(2) Time of ascent = time of descent = u/g.
(3) Total time of flight = 2u/g.
(4) Velocity of fall at the point of projection = u (downwards).
(5) Gallileo's law of odd numbers : For a freely falling body released from rest ratio of successive distance covered in equal time interval 't'.
S₁ : S₂ : S₃ : . . . . . Sₙ = 1 : 3 : 5 : . . . . . : 2n - 1.
(6) At any point on its path the body will have same speed for upward journey and downward journey. if a body thrown upwards crosses a point in time t₁ & t₂ respectively then height of point h = 1/2gt₁t₂.
(7) Maximum height H = 1/8 g(t₁ + t₂)².
(8) A body is thrown upward, downward & horizontally with same speed take times t₁, t₂ & t₃ respectively to reach the ground then t₃ = √t₁t₂ & height from where the particle was thrown is H = 1/2gt₁t₂.