# Answer to Question #5477 in Electric Circuits for jessica

Question #5477

a base ball is thrown straight upward on the Moon with an initial speed of 35m/s^-1.Compute a) the maximum height reached by the ball b) the time taken to reach that height c) its velocity 10s after it is thrown and d) when the balls height is 50m.

Expert's answer

a base ball is thrown straight upward on the Moon with an initial speed of V0 = 35m/s. Compute

// g = 1.622 m/s^2 (Moon gravity)

b) the time taken to reach that height

V0 = g*T ==> T = V0/g = 35/1.622 ≈ 21.5783 s.

a) the maximum height reached by the ball

V = 35;

g=1.622;

T=21.5783;

H = V0*T/2 = ( V0*T - (g*T^2)/2 ) = 35*21.5783/2 = 377.6203 m.

c) its velocity 10s after it is thrown

V = V0 - g*10 = 35 - 1.622*10 = 18.7800 m/s.

d) when the balls height is 50m

H = V0*T - (g*T^2)/2

(g/2)*T^2 - V0*T + H = 0

T1 = [V0 - sqrt(V0^2 - 4*(g/2)*H)] / 2*(g/2) = [V0 - sqrt(V0^2 - 2*g*H)] / g = [35 - sqrt(35^2 - 2*1.622*50)] / 1.622 ≈ 1.4811 s

T2 = [V0 + sqrt(V0^2 - 4*(g/2)*H)] / 2*(g/2) = [V0 + sqrt(V0^2 - 2*g*H)] / g = [35 + sqrt(35^2 - 2*1.622*50)] / 1.622 ≈ 41.7288 s

That means that the ball reaches the height 50 m two times: once while moving up and once while falling down.

// g = 1.622 m/s^2 (Moon gravity)

b) the time taken to reach that height

V0 = g*T ==> T = V0/g = 35/1.622 ≈ 21.5783 s.

a) the maximum height reached by the ball

V = 35;

g=1.622;

T=21.5783;

H = V0*T/2 = ( V0*T - (g*T^2)/2 ) = 35*21.5783/2 = 377.6203 m.

c) its velocity 10s after it is thrown

V = V0 - g*10 = 35 - 1.622*10 = 18.7800 m/s.

d) when the balls height is 50m

H = V0*T - (g*T^2)/2

(g/2)*T^2 - V0*T + H = 0

T1 = [V0 - sqrt(V0^2 - 4*(g/2)*H)] / 2*(g/2) = [V0 - sqrt(V0^2 - 2*g*H)] / g = [35 - sqrt(35^2 - 2*1.622*50)] / 1.622 ≈ 1.4811 s

T2 = [V0 + sqrt(V0^2 - 4*(g/2)*H)] / 2*(g/2) = [V0 + sqrt(V0^2 - 2*g*H)] / g = [35 + sqrt(35^2 - 2*1.622*50)] / 1.622 ≈ 41.7288 s

That means that the ball reaches the height 50 m two times: once while moving up and once while falling down.

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