Question

A body is projected upwards from level ground such that its velocity makes an angle of 45 with the horizontal at half at max height. The initial angle of projection with the horizontal is tan^-1,where x is

Answer 1

Explanation:

Given that,

Given that,Angle θ=45°

Horizontal distance x=10m

Horizontal distance x=10mVertically distance y=7.5m

Horizontal distance x=10mVertically distance y=7.5mAcceleration due to gravity g=10m/s2

Horizontal distance x=10mVertically distance y=7.5mAcceleration due to gravity g=10m/s2

Horizontal distance x=10mVertically distance y=7.5mAcceleration due to gravity g=10m/s2 Now,

Horizontal distance x=10mVertically distance y=7.5mAcceleration due to gravity g=10m/s2 Now,y=(tanθ)x−( g/2u 2cos2θ)x 2

)x 2

)x 2 7.5=1×10− 10/(2u 2×1/2)100

7.5−10=− 1000/u2

2

u 2= 1000/2.5

u=20m/s

u=20m/sHence, the initial speed of the projection is 20 m/s