Question #11461

The ceiling of a long hall is 25m high.What is the maximum horizontal distance that a ball thrown with a speed of 40ms-1 can go without hitting the ceiling of the hall ?

Expert's answer

hmax = 25 m

hmax = V0y * t - g * t^2 / 2, t = V0y / g, V0y = V0 *

sinA

hmax = V0^2 * (sinA)^2 / g - (g / 2) * V0^2 * (sinA)^2 / g^2

hmax =

V0^2 * (sinA)^2 / g - V0^2 * (sinA)^2 / (2 * g) = V0^2 *

(sinA)^2 / (2 *

g)

25 = (sinA)^2 * 40^2 / (9.81 * 2)

sinA = sqrt(25 * 9.81 * 2 / 1600) =

0.5537

A = 33.62 degrees

lmax = V0x * t, t = V0y / g, V0x = V0 * cosA,

V0y = V0 * sinA

lmax = V0 * cosA * V0 * sinA / g = 40 * 40 * 0.8327 * 0.5537

/ 9.81 =

[approx.] = 75 m

Answer: 75 m.

hmax = V0y * t - g * t^2 / 2, t = V0y / g, V0y = V0 *

sinA

hmax = V0^2 * (sinA)^2 / g - (g / 2) * V0^2 * (sinA)^2 / g^2

hmax =

V0^2 * (sinA)^2 / g - V0^2 * (sinA)^2 / (2 * g) = V0^2 *

(sinA)^2 / (2 *

g)

25 = (sinA)^2 * 40^2 / (9.81 * 2)

sinA = sqrt(25 * 9.81 * 2 / 1600) =

0.5537

A = 33.62 degrees

lmax = V0x * t, t = V0y / g, V0x = V0 * cosA,

V0y = V0 * sinA

lmax = V0 * cosA * V0 * sinA / g = 40 * 40 * 0.8327 * 0.5537

/ 9.81 =

[approx.] = 75 m

Answer: 75 m.

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