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
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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