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The height y and the horizontal distance x for a projectile at any instant t after it is projected at an angle theta with horizontal are given by eq. 2y = 16t - 10t^(2) and x = 6t metre when t is in sec. the velocity with which the projectile projected is : |
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Answer» `6ms^(-1)` `implies ucostheta=6`…(i) and `2y=16t - 10T^(2)implies y=8t-5t^(2)` or `usintheta t-1/2g t=8t-5t^(2)` `impliesusintheta=8`…(II) Squaring and adding (i) and (ii) gives u=10m/s |
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