Asked by Swasthi
A stone is thrown from the top of a tower of height 60m with velocity 40m/s at an angle of 30 degree above the horizontal. Calculatethe time taken by the stone to hit the ground. At what distance from the foot of the tower the stone hits the ground.
Answers
Answered by
Henry
Vo = 40m/s[30o].
Xo = 40*Cos30 = 34.64 m/s.
Yo = 40*sin30 = 20 m/s.
Y = Yo + g*Tr.
0 = 20 - 9.8Tr, Tr = 2.04 s.
ha = Yo*Tr + 0.5g*Tr^2.
ha = 20*2.04 - 4.9*2.04^2 = 20.4 m. = Ht. above tower.
h = 20.4 + 60 = 80.4 m. Above gnd.
0.5g*Tf^2 = 80.4.
4.9Tf^2 = 80.4, Tf = 4.05 s. = Fall time.
a. Tr+Tf = 2.04 + 4.05 = 6.09 s. To hit gnd.
b. D = Xo*(Tr+Tf) = 34.64 * 6.09 = 211 m.
Xo = 40*Cos30 = 34.64 m/s.
Yo = 40*sin30 = 20 m/s.
Y = Yo + g*Tr.
0 = 20 - 9.8Tr, Tr = 2.04 s.
ha = Yo*Tr + 0.5g*Tr^2.
ha = 20*2.04 - 4.9*2.04^2 = 20.4 m. = Ht. above tower.
h = 20.4 + 60 = 80.4 m. Above gnd.
0.5g*Tf^2 = 80.4.
4.9Tf^2 = 80.4, Tf = 4.05 s. = Fall time.
a. Tr+Tf = 2.04 + 4.05 = 6.09 s. To hit gnd.
b. D = Xo*(Tr+Tf) = 34.64 * 6.09 = 211 m.
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