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A ball thrown with speed of 12m/s at an angle of $60^\circ $with a building strikes the ground 11.3 m horizontally from the foot of the building as shown. Determine the height of the building.
written 7.4 years ago by gravatar for teamques10 teamques10 ★ 65k modified 2.3 years ago by gravatar for pedsangini276 pedsangini276 • 4.8k

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : DEC 2015
engineering mechanics
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written 7.4 years ago by gravatar for teamques10 teamques10 ★ 65k

Horizontal ( +ve) Vertical ( +ve)

Initial velocity $\hspace{1cm}V_H=12 \sin 60 m/s \hspace{1cm} u_V=12 \cos60 m/s$

Acceleration $\hspace{1cm}a_H= 0 \hspace{1cm} a_V=g=9.81m/s^2$

Displacement $\hspace{1cm} x=11.3m \hspace{1cm} y=H$

Final velocity $\hspace{1cm} V_H=12 \sin 60 m/s\hspace{1cm} v_V=?$

Time $\hspace{1cm}t=? \hspace{1cm}t=?$

Horizontal motion:-

Using $s=ut+ \dfrac 12 at^2,$ we get,

$\hspace{1cm} x=V_H t [\because …

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