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For the top of a tower 156.8 m high, a ...

For the top of a tower ` 156.8 m` high, a projectile is thrown up with a velcity of `39.2 ms^(-1)`, makingan angle `30%(@)` with borizontal direction. Find the distance from the foot of tower wher it strikes the ground and the time taken byit do so.

Text Solution

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The situation is shown Here height of tower
OA = 156.8m
`u=39.2ms^(-1)`
`theta=30^(@)`
time for which projectile remain is air = t = ?

Horizontal distance covered R = OD = ?
Now `u_x=ucos theta` and `u_y=usin theta` be the components of velocity `baru`
Motion of projectile from O to H to D
Using equation y`=u_yt+1/2a_yt^2`
here : `y=15.9m,u_y=-usintheta`
`39.2sin30^(@)`
`a_y=9.8m//s^2,t=?`
`156.8=-39.2xx0.5t+4.9t^2`
`156.8=-19.6t+4.9t^2`
or `4.9t^2-19.6t-156.8=0`
or `t^2-4t-19.6t-156.8=0`
or `t^2-4t-32=0implies(t-8)(t+4)=0`
We get t=8s , t=-4s
t = – 4 s is not possible, thus we take t = 8s. Now horizontal distance covered in this time
`R=u_xxxt=ucosthetaxxt=39.2xxcos30^(@)xxt`
R=271.57m
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