Question

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the hortzontal and with an Initilal speed of 19.0 m/s. The point of release is PRACTICE IT 49.0 m above the ground. (a) How long does it take for the ball to hit the ground? (b) Find the ball's speed at impact. m/s (e) Find the horizontal range of the ball. EXERCISE HINTS: GETTING STARTEDPM STUCK Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0° below the horizontal. If it strikes the ground 52.2 m away, find the following. (Hint: For part ( equation for the r-displacement to eliminate vot from the equation for the y-displacement.) a), use the m/s (e) the speed and angle of the velocity vector with respect to the horizontal at impact speed e below the horizontal

Answer 1

Ans:-

Given data:-

Θ= 30deg Vi= 19m/s, h= 49m

Vix= cos30 *19= 16.45m/s

Viy= sin30*19 = - 9.5m/s( downward direction)

a]here we use kinematical equation along Y direction

Y- Y0 = Viy*t + ½ *at^2

49 = -9.5 t + ½ 9.8 *t^2

4.9t^2 - 9.5t – 49 = 0

After solving we get

t= 4.28s

b] Vf = ?

Vfy^2 = Viy^2 + 2ays

         = (-9.5)^2 + 2*(9.8)*49

Vfy^2 = 90.25 + 960.4 = 1050.65

Vfy = 32.41m/s

C] R =( Vix^2/g)*sin2θ

R = (16.45^2/9.8)sin(2*30)

R = 23.91m