Question
The height $y$ and the distance $x$ along the horizontal planc of a projectile on a certain planct (assuming flat surfacc) with no surrounding atmosphere are given by $x=6 t$ and $y=8 t-5 t^{2}$, wherc $x$ and $v$ are in metre and time $t$ is in second. Find the velocity with which the body is projected, the maximum height attained and the range of the projectile. Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$
Step 1
This gives us the velocity as a function of time: \begin{align*} v = \frac{dy}{dt} = 8 - 10t \end{align*} At $t=0$, the velocity is the initial velocity $u$, so $u = 8 - 10 \cdot 0 = 8 \, \text{m/s}$. Show more…
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The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planct (with no surrounding atmosphere) are given by $y=\left(8 t-5 t^{2}\right)$ motre and $x=6 t$ meter, where $t$ is in second The velocity with which the projectile is projected is (a) $8 \mathrm{~m} / \mathrm{s}$ (b) $6 \mathrm{~m} / \mathrm{s}$ (c) $10 \mathrm{~m} / \mathrm{s}$ (d) insulficient data
Motions in Two and Three Dimensions
Section B
The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y=8t-5t^2 meter and x=6t meter, where t is in second. The velocity with which the projectile is projected is ,find resultant velocity
The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=\left(8 t-5 t^{2}\right) \mathrm{m}$ and $x=6 t$ $\mathrm{m}$, where $t$ is in seconds. The velocity with which the projectile is projected at $t=0$, isa. $8 \mathrm{~m} / \mathrm{s}$ b. $6 \mathrm{~m} / \mathrm{s}$ c. $10 \mathrm{~m} / \mathrm{s}$ d. not obtainable from the data
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