A toy rocket is launched from the ground with an initial speed of 40 m/s . If the maximum height reached by the rocket is 40 m , what is its angle of projection with the horizontal? (take g = 10 m/s ^2 )

Step-by-step solution

Given: initial speed u = 40 m/s , maximum height H = 40 m , and g = 10 m/s ^2 . The formula for the maximum height of a projectile is: H = u^2 ^2 2g Substitute the known values into the equation: 40 = (40)^2 ^2 2 10 40 = 1600 ^2 20 40 = 80 ^2 Solving for ^2 : ^2 = 40 80 = 1 2 Taking the square root of both sides: = 1 2 The angle whose sine is 1 2 is 45^ . Thus, the angle of projection is 45^ . Answer: 45^