A baseball player at second base throws a ball 90 feet to the player at first ba

Question

A baseball player at second base throws a ball 90 feet to the player at first base. The ball is released at a point 5 feet above the ground with an initial velocity of 50 miles per hour and at an angle of 15 degree above the horizontal. At what height does the player at first base catch the ball? Take the acceleration due to gravity to be a constant 32 feet per second squared and ignore air resistance. Hints: there are 3600 seconds in 1 hour, there are 5280 feet in one mile, and see 12.3 example 6 for a similar problem.

Explanation / Answer

r(t)= vo cos t i + (h+vo sin t -16t2)j

length component l(t) =  vo cos t

height component h(t) = h+vo sin t -16t2

given data

= 15 degrees , h = 5 ft, initial velocity = 50mph

v0 = 50 mph = (50x5280)/3600 = 73.3 ft/s

now, r(t) = 73.3 cos15 t i + (5+73.3 sin 15t -16t2)j

when l = 90ft = 73.3 cos15 t

==> t = 1.27 sec

now, reqd. height

= 5 + 73.3sin15 (1.27- 16 (1.27)2 = 3.287 ft

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