*EXERCISE Suppose the ball is thrown from the same height as in the PRACTICE IT

Question

*EXERCISE Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 31.0°below the horizontal. If it strikes the ground 44.6 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)

(a) the time of flight_______s

(b) the initial speed_______m/s

(c) the speed and angle of the velocity vector with respect to the

horizontal at impact speed_______m/s angle________° below the horizontal

Explanation / Answer

Let initial speed be U m/s and time of flight be T seconds.

Horizontal component velocity = Ucos = Ucos31
Initial Vertical component velcoity = Usin = Usin31

Height = This is not clear from your problem. Please mention it. Suppose this is 'h'.

And Range = 44.6m .

X = Ucos31 x T
44.6 = Ucos31 x T
From the above two expressions, you can find the value of T in the form of U.

h = ½(9.8)(T)² + Usin31xT

Now,

h = ½(9.8)(44.6/Ucos31)² + Usin31/(44.6/Ucos31)

(a) & (b) Put the value of 'h', you can find out U and T.

Vertical componenet at impact = Usin31 + (T)(9.8)

Speed at impact = (Ucos31)² + (Verticle component of impact)²
(c) Angle =Arctan (Verticle component of impact)/Ucos31

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