Question

A 2.0 kg ball is thrown upward with an initial speed of 24.0 m/s from the edge of a 46.0 m high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 5.80 m/s. The woman runs in a straight line on level ground. Ignore air resistance on the ball. How far does she run before she catches the ball? (Your result must be in units of meters and include one digit after the decimal point. Maximum of 3% of error is accepted in your answer. Take g=9.80 m/s2.)

Answer #1

answer) from the question and by applying law of kinematics the condition at which she catches the ball is

horizontal speed ( in x direction)=woman's speed

v1=v2cos

=cos-1(v1/v2)=cos-1(5.80/24.0)=76o

now applying projectile range formula we have

R=v2cos/g*(v2sin+(v2sin)2-2gyR=24*cos76/9.8*(24*sin76+(24sin76)2+2*9.8*46 m)=36.3 m

so answer is 36.3 m

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