Question #88f6b

1 Answer
yumean1119
Dec 10, 2017

The ball will clear the wall only when the initial speed is more than about 19 m/s.

Explanation:

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Let
#v_0# (m/s): initial speed of the ball
#v_(0x)# (m/s): initial speed along the horizontal direction
#v_(0y)# (m/s): initial speed along the vertical direction.

Then, it will take #t=32/v_(0x)=32/(v_0cos40°)=41.77/v_0# (s) to reach the wall.
At that time, the height of the ball is
#h=v_(0y)t-1/2g t^2#
#=v_0sin40° * 41.77/v_0 -1/2 * 9.807 * (41.77/v_0)^2#
#=26.85-8555/v_0^2# [m].

When the ball clears the wall, the inequation
#26.85-8555/v_0^2>3.5# must be satisfied.
#8555/v_0^2<23.35#
#8555<23.35v_0^2#
#v_0^2> 366.4#
#v_0>19.14# (m/s)

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