stant Part A A baseball thrown at an angle of 55.0 above the horizontal strikes

Question

stant Part A A baseball thrown at an angle of 55.0 above the horizontal strikes a building 18.0 m away at a point 5.00 m above the point from which it is thrown. Ignore air resistance. Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). m/s Submit Part B Find the magnitude of the velocity of the baseball just before it strikes the building u= m/s Submit Request Answer Part C Find the direction of the velocity of the baseball just before it strikes the building

Explanation / Answer

Here ,

theta = 55 degree

x = 18 m

y = 5 m

part a) let the initial velocity is u

Using the equation of trajectory

y = x * tan(theta) - g * x^2/(2 * (u * cos(theta))^2)

5 = 18 * tan(55 degree) - 9.8 * 18^2/(2 * (u * cos(55 dgeree))^2)

solving for u

u = 15.3 m/s

the magnitude of initial speed is 15.3 m/s

part B)

at the building , let the speed is v

v^2 - 15.3^2 = - 2 * 9.8 * 5

v = 11.7 m/s

the magnitude of velocity is 11.7 m/s

b) as the horizontal component of velocity is always constant

15.3 * cos(55 degree) = 11.7 * cos(theta)

theta = 41.4 degree

the direction is 41.4 degree above horizontal

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