Question

Projectile is projected with a velocity 10m/s at an angle 37 degree to the horizontal.find the location at which the particle is at a height of 1m from the point of projection

Answer 1

Answer:

solved

Explanation:

x = 10tcos37

y = 10tsin37 - 5t² , g = 10 m/s²

t = x/10cos37

y = xtan37 - 0.05x²sec²37 = 1

0.05x²sec²37  - xtan37 +1 = 0

x² - 10xsin74+ 20cos²37 = 0

x = {10sin74±√(100sin²74 - 80cos²37)} ÷2 = 5sin74± 5sin74√1 - 0.2csc²37

x = 5sin74(1±√1 - 0.2csc²37)

x = 5sin74(1±√1 - 0.2csc²37) = 5sin74(1±0.67)

x1 = 5*1.67sin74 = 8.03 m.

x2 = 5*0.33sin74 = 1.59 m.