(a) (i) During a test match at The Oval, a cricket ball is thrown back to the wi

Question

(a) (i) During a test match at The Oval, a cricket ball is thrown back to the wicket keeper,
upward of the horizontal at an angle of 15, from a throwing height of 1.5 m. If it is
released at a speed of 20 m s??1, how far will the ball have traveled horizontally if the
wicket keeper misses catching it entirely, and it hits the ground? You may ignore air
resistance. [4 marks]
(ii) If instead the ball is thrown downwards at an angle to the horizontal of 15, how far will
it travel, horizontally, before it reaches the ground. Initial speed and throwing height are
unchanged. [1 mark]

Explanation / Answer

vertical component of velocity ,Vy=15sin(15)=3.88m/s

horizontal component of velocity ,Vx=15cos(15)=14.489m/s

since s=ut+(1/2)at^2

applying equation of motion in y direction

here s=-1.5 , u=+3.88 m/s , a=-9.8 m/s^2

-1.5=3.88t-0.5*9.8*t^2

solving the equation we get;

t=1.076 s [other value of t is not possible]

ball remains in air for this amount of time

so horizontal distance travalled by ball = (horizontal velocity)*(time of flight)

= 14.489*1.076 m

=15.59 m

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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