A particle is projected from a tower as shown in figure, then the distance from the foot of the tower where it will strike the ground will be – (take g = 10 m/s² and sin 37^o =0.6)

(a)    4000/3 m

(b)    5000/3 m

(c)    2000 m

(d)    3000 m

Question

A particle is projected from a tower as shown in figure, then the distance from the foot of the tower where it will strike the ground will be – (take g = 10 m/s² and sin 37^o =0.6)

(a) 4000/3 m

(b) 5000/3 m

(d) 3000 m

Simran Bhatia · Answer

(a) &ndash;1500 = &ndash;100t &ndash; 1/2 &times; 10t2 t^2/2 + 10t &ndash; 150 = 0 => t = (-20&plusmn;40)/2 So, t = 10 sec. i.e., horizontal distance = 500/3&times;4/3&times;10=4000/3 m

Bipasha Kalita · Answer

y=Uy - 1/2 gt^2. -1500=500/3×sin37°t-5t^2 Solving for t we get, t=10 Range,D=Ux t. =500/3 × cos 37°×10 =4000/3m

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A particle is projected from a tower as shown in figure, then the distance from the foot of the tower where it will strike the ground will be – (take g = 10 m/s 2 and sin 37 o = 0.6) (a) 4000/3 m (b) 5000/3 m (c) 2000 m (d) 3000 m

A particle is projected from a tower as shown in figure, then the distance from the foot of the tower where it will strike the ground will be – (take g = 10 m/s² and sin 37^o =0.6)

(a)    4000/3 m

(b)    5000/3 m

(c)    2000 m

(d)    3000 m

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