A ball is tossed from an upper-story window of a building. The ballis given an i

Question

A ball is tossed from an upper-story window of a building. The ballis given an initial velocity of 7.60 m/sat an angle of 24.0° below the horizontal. It strikes the ground5.00 s later. (a) How far horizontally from the base of thebuilding does the ball strike the ground?
1 m

(b) Find the height from which the ball was thrown.
2 m

(c) How long does it take the ball to reach a point 10.0 m belowthe level of launching?
3 s
Any help would be greatly appreciated i am so lost and dont knowwhere to start (a) How far horizontally from the base of thebuilding does the ball strike the ground?
1 m

(b) Find the height from which the ball was thrown.
2 m

(c) How long does it take the ball to reach a point 10.0 m belowthe level of launching?
3 s

Explanation / Answer

   Horizontal component of initialvelocity      ux   =   u* cos    =   7.60 * cos240   =   6.94   m/s    Vertical component of initialvelocity         uy   =   u* sin    =   7.60 * sin240   =   3.09   m/s    a.   Horizontaldistance   x   =   ux* t   =   6.94 *5.00   =   34.70   m,   fromthe base    b.   Height ofbuilding      h   =   uy* t   +   (1/2) * g *t2          h   =   3.09* 5.00   +   0.5 * 9.8 *5.002   =   137.95   m    c.   Second equation is             h   =   uy* t   +   (1/2) * g *t2          10   =   3.09* t   +   0.5 * 9.8 * t2          4.9 *t2   +   3.09 *t   -   10   =   0          This isa quadratic equation in t, ofform   ax2   +   bx   +   c   =   0          thesolution is             x   =   {-b±(b2   -   4ac)}/2a          Substituting   b   =   3.09,   a   =   4.9,   c   =   -10          t   =   +1.15   s         or      t   =   -1.77   s       since time can not benegative, hence the ball would take 1.15 s to descend to 10.0 mbelow the level of lanching.          4.9 *t2   +   3.09 *t   -   10   =   0          This isa quadratic equation in t, ofform   ax2   +   bx   +   c   =   0          thesolution is             x   =   {-b±(b2   -   4ac)}/2a          Substituting   b   =   3.09,   a   =   4.9,   c   =   -10          t   =   +1.15   s         or      t   =   -1.77   s       since time can not benegative, hence the ball would take 1.15 s to descend to 10.0 mbelow the level of lanching.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.