The equation of trajectory of a projectile thrown from a point on the ground is y=(x−x2/40)m. If g=10ms−2, the maximum height reached is

# The equation of trajectory of a projectile thrown from a point on the ground is $y=\left(x-{x}^{2}/40\right)m$. If $g=10m{s}^{-2}$, the maximum height reached is

1. A

6 m

2. B

8 m

3. C

10 m

4. D

12 m

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$A=tan\theta =1$

$⇒\theta =45º$

$\frac{g}{{u}^{2}{\mathrm{cos}}^{2}\theta }=\frac{1}{40}$

Maximum height (H)

$=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta }{2g}=\frac{20×20}{2×10}×{\left(\frac{1}{\sqrt{2}}\right)}^{2}=10m$

## Related content

 Zener Diode Hygrometer Half Wave Rectifier Compound Microscope Solar Cooker MCB Difference Between Scalar and Vector Quantity Difference Between Interference and Diffraction Difference Between Heat and Temperature Difference Between Real and Virtual Image

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)