A ball s projected from a point on horizontal ground. The speed of projection is 30 m/s and the greatest height reached is 20 meter. Assuming no air resistance, find the angle of projection above the horizontal ?

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Answer 1 · 2017-06-28T08:05:10

The angle is #=41.3º#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u=30sintheta#

The height is #h=s=20m#

The acceleration is #a=(-g)ms^-2#

The velocity, at the greatest height is #v=0ms^-1#

We apply, the equation of motion

#v^2=u^2+2as#

#0=(30sintheta)^2-2*g*20#

#(30sintheta)^2=40g#

#30sintheta=sqrt(40g)#

#sintheta=sqrt(40g)/30=0.66#

#theta=41.3º#