Question

A rock is thrown into the air with an initial velocity of 40 m/s and 60 degrees above the horizontal. (a) What is the time when the velocity makes 45 degrees above the horizontal? (b) What is the time when the velocity makes 45 degrees below the horizontal? (c) Does it exist when the velocity makes zero degrees with the horizontal? If yes, when? (d) Does it exist when the velocity makes 90 (or -90) degrees with the horizontal? If yes, when?

Answer 1

u ( initial speed) = 40 m/s

initial projection angle = 60 degrees

A) final angle = 45 degree

using formula,

Tan(f) = V_y/V_x

Tan(f) = [USin - gt]/UCos

Tan(45) = [40Sin60 - 9.8t]/40Cos(60)

1 = [40Sin60 - 9.8t]/20

t = 1.5 seconds.

B)

final angle = -45 degree

using formula,

Tan(f) = V_y/V_x

Tan(f) = [USin - gt]/UCos

Tan(-45) = [40Sin60 - 9.8t]/40Cos(60)

-1 = [40Sin60 - 9.8t]/20

t = 5.58 seconds.

C)

final angle = 0 degree

using formula,

Tan(f) = V_y/V_x

Tan(f) = [USin - gt]/UCos

Tan(0) = [40Sin60 - 9.8t]/40Cos(60)

0 = [40Sin60 - 9.8t]/20

t = 3.53 seconds.

D)

final angle = 90 degree

using formula,

Tan(f) = V_y/V_x

Tan(f) = [USin - gt]/UCos

Tan(90) = [40Sin60 - 9.8t]/40Cos(60)

infinity = [40Sin60 - 9.8t]/20

t = infinity .

Therefore, it is not possible for the rock to make an angle of 90 degree at any time with the horizontal.

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