Question

A ball is thrown from the top of a building at an angle of 30degrees above the horizontal and with an initial speed of 20m/s. if the ball is in flight for 4seconds A) how tall is the building? b)what horizontal distance does the ball travel? c) what maximum height does the ball reach? d)with what speed and angle of impact does the ball land?

Answer 1

(a)

along vertical

voy = vo*sintheta

acceleration ay = -g = -9.8 m/s^2

from equation of motion

y-Y0 = voy*T + 0.5*ay*T^2

0-H = (20*sin30*4) - (0.5*9.8*4^2)

Height H = 38.4 m

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(b)

along horizantal

initial velocity vox = vo*costheta

ax = 0

from equaiton of motion

x-Xo = vox*T+ 0.5*ax*T^2

x-X0 = vo*costheta*T

X - 0 = 20*cos30*4

X = 69.3 m <<<===ANSWER

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(c)

at maximum height final speed vy = 0

vy^2 - voy^2 = 2*ay*(y-y0)

0 - (20*sin30)^2 = -2*9.8*(y-38.4)

y = 43.5 from the ground <<<=======answer

++++++++++++++++++++

(d)

along vertical after T time

vy = voy + ay*T

vy = (20*sin30) - (9.8*4) = -29.2 m/s

along orizantal

vx = vox = 20*cos30 = 17.32 m/s

speed = sqrt(vx^2+vy^)

speed = sqrt(29.2^2+17.32^2) = 33.9 m/s <<<=====answer

direction = tan^-1(Vy/Vx)

direction = -59.32   <<<<<=====answer