Physics · Motion in a plane, projectile motion, uniform circular motion
A ball is thrown from the top of a tower with an initial velocityof at an angle
A ball is thrown from the top of a tower with an initial velocityof \( 10 \mathrm{m} / \mathrm{s} \) at an angle of \( 30^{\circ} \) above the horizontal. It hits theground at a distance of \( 17.3 \mathrm{m} \) from the base of the tower.The height of the tower \( \left(\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}\right) \) will be
- A. \( 10 \mathrm{m} \)
- B. \( 110 \mathrm{m} \)
- C. 12 \mathrm{m} \)
- D. \( 100 \mathrm{m} \)
Step-by-step solution
Given initial speed u=10 m/s at 30°, horizontal component u_x=10 cos30=5√3 m/s, vertical component u_y=10 sin30=5 m/s. Horizontal range R=17.3 m = 10√3 m, so time of flight T=R/u_x = (10√3)/(5√3)=2 s. Using vertical motion: s_y = u_y t + (1/2) a_y t^2, with a_y=-g=-10 m/s^2, s_y = 5*2 + 0.5*(-10)*4 = 10 -20 = -10 m. The negative sign indicates displacement downward, so tower height = 10 m.
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