Physics, asked by irshadkhan54455, 2 months ago

In a cricket match, a batsman hits a ball at an angle of 36o from the horizontal with an initial speed of 15.5 m/s. Assuming that the ball moves in a vertical plane, find (a) the time at which the ball reaches the highest point of its trajectory, (b) its maximum height, (c) its time of flight and range, and (d) its velocity when it strikes the ground.

Answers

Answered by amitnrw
1

Given : a batsman hits a ball at an angle of 36° from the horizontal with an initial speed of 15.5 m/s.

To Find : (a) the time at which the ball reaches the highest point of its  trajectory,

(b) its maximum height,

(c) its time of flight and range, and

(d) its velocity when it strikes the ground.

Solution:

 36° from the horizontal with an initial speed of 15.5 m/s.

Horizontal Speed =  15.5 cos36°   =  12.54 m/s

Vertical Speed =  15.5 Sin36°   =   9.11  m/s

Vertical Speed  = 0 when it  reaches the highest point of its trajectory

Using V = U + at

a = -g = - 9.8 m/s²

=> 0  =  9.11 + (-9.8)t

=> t = 0.93sec

time at which the ball reaches the highest point of its trajectory = 0.93 sec

Time of flight = 2 * 0.93  = 1.86 sec

range  = Horizontal Speed  *Time of flight

=  12.54 * 1.86

= 23.324 m

V² - U² = 2aS

=>  - 9.11² = 2(-9.8)S

=> S = 4.234

maximum height = 4.234  m

KE + PE initiall = KE + PE  Final

PE = 0 initially and finally

Hence KE initially and finally are equal

=> velocity is same as initially

velocity when it strikes the ground. = 15.5 m/s

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