A ball is thrown from the top of a building with an initial velocity of 21.4 m/s

Question

A ball is thrown from the top of a building with an initial velocity of 21.4 m/s straight upward, at an initial height of 55.3 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure. (a) Determine the time needed for the ball to reach its maximum height. s (b) Determine the maximum height. m (c) Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. Time s Velocity m/s (d) Determine the time needed for the ball to reach the ground. s (e) Determine the velocity and position of the ball at t = 5.42 s. Velocity m/s Position m

Explanation / Answer

A) apply V= U+at => T = U/a = 21.4/9.8 = 2.183 sec Answer

B) apply V^2 - U^2 = 2*a*s => h = 21.4*21.4/(2*9.8) = 23.365 m Answer

C) Time = 2*2.183 = 4.366 sec Answer

V = -21.4 m/s = 21.4 m/s downwards Answer

D) apply s = ut + 0.5*a*t^2

=> 4.9t^2 - 21.4t = 55.3 => t = 6.19 sec Answer

E) V = 21.4 - 9.8*5.42 = -31.716 m/s = 31.716 m/s downwards

s= 21.4*5.42 - 4.9*5.42*5.42 = -27.956 = 27.956 m below from top of buliding

so 55.3 - -27.956 = 27.344 m above the groiund Answer

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