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In a criket match a batsman hits a six . Ball is hit 1 m above the ground level from a point which is 110m away from the benches of stadium . Ball leaves the bat with a speed 35 m//s and at an angle 53^(@) abovethe horizontal. Height and width of benches are 1 m each . Assume thatbenches are perpendicular to the plane of motion of the ball. Which bench will the ball hit ? |
Answer» Solution :Situation DESCRIBED in question is shown inthe following figure . Initial point of motion of ball is MARKED as point O and is selectedas the origin of the coordinate system . X-axis is horizontal and Y-axis is vertically upward.  Benches are 1 m high and 1 m wide and if we join the edge of the benches, then we get a planeinclined at an angle `45^(@)` with the horizontal . On thisimaginary palne , length of each step is `sqrt2m` . Let the ball hit on thisplane at a distnce x from its bottom . Components of displacement of ball can be written as follows : `S_(x)=110+x cos 45^(@)=110+(x)/(sqrt2),` `S_(y)x sin 45^(@)=(x)/(sqrt2),` Components of initial velocity are shown in the figure . `u_(x)=21m//s,u_(y)=28m//s,` We can write the components of acceleration as follows: `a_(x)=0,a_(y)=-9.8m//s^(2)`, X-axis, `S_(x)=u_(x)t+(1)/(2)u_(x)t^(2) rArr 110+(x)/(sqrt2)=21t` Y-axis : `S_(y)=u_(y)t+(1)/(2)a_(y)t^(2) rArr(x)/(sqrt2)=28t-4.9t^(2)` Substituting VALUE of `(x)/(sqrt2)` from equation (ii) in equation (i), we get the following: `rArr 110+28t-4.9t^(2)=21t` `rArr 4.9t^(2)-7t-110=0` `rArr t=(7pmsqrt(49+4xx4.9xx110))/(9.8)` `rArr t=(7 pm 7sqrt45)/(9.8)` We shall consider the positive sigh only and rearrange as follows: `rArr t=(1+3sqrt5)/(1.4)=5.5s` Substituting above result in equation (1), `rArr 110 +(x)/(sqrt2)=21xx5.5 rArr x=5.5 sqrt2` We have already discussed that each step is `sqrt2` m, HENCE `(x)/(sqrt2)=5.5` We can that it is more than `5^(t"th")` step , hence ball hits the `6^("th")` step. |
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