Question

What is a projectile?Derive the expressions for the (i) time of flight (ii) maximum height (iii) horizontal range of aprojectile projected with a speed v making an angle ɵwith the horizontal direction from the ground. Abody is projected with a velocity of 40 m/s. After2s it crosses a vertical pole of height 20.4m.What is its angle of projection?

Answer 1

Answer:

1. an object, such as a bullet, that is fired from a gun or other weapon.
2. time of flight-->T=2visinθg T = 2 v i sin ⁡ . The angle of reach is the angle the object must be launched at in order to achieve a specific distance: θ=12sin−1(gdv2) θ = 1 2 sin − 1 ⁡ ( gd v 2 ) .
3. maximum height--->yo = 0, and, when the projectile is at themaximum height, vy = 0. Note that themaximum height is determined solely by the initial velocity in the y direction and the acceleration due to gravity.
4. horizontal range of aprojectile projected with a speed --->The range (R) of the projectile is thehorizontal distance it travels during the motion. Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the verticaldirection is usina (by resolving). Hence: y = utsina
5. -
6. ANSWER
8. ANSWERx=40cosθ×2=80cosθ
10. ANSWERx=40cosθ×2=80cosθApplying equation of trajectory ,
12. ANSWERx=40cosθ×2=80cosθApplying equation of trajectory ,20=80cosθ×tanθ−2×cos2θ×402g(80cosθ)2
14. ANSWERx=40cosθ×2=80cosθApplying equation of trajectory ,20=80cosθ×tanθ−2×cos2θ×402g(80cosθ)220=80sinθ−240
16. ANSWERx=40cosθ×2=80cosθApplying equation of trajectory ,20=80cosθ×tanθ−2×cos2θ×402g(80cosθ)220=80sinθ−240sinθ=21,θ=30∘
18. ANSWERx=40cosθ×2=80cosθApplying equation of trajectory ,20=80cosθ×tanθ−2×cos2θ×402g(80cosθ)220=80sinθ−240sinθ=21,θ=30∘Range =gu2sin2θ=201600×3=803=138.5m

Explanation:

Hope it will help you ☆☆☆