Answer to Question #292342 in Mechanics | Relativity for Nash

Question

An object is being shot from a horizontal ground at an incline angle of 30 degree with respect to the ground at a speed of 46 m/s. Find the duration in seconds that the object is above the height of 19 m. Give your answer with one decimal place.

Answer 1

Answer

This is vertical speed component

u=46sin 30°=23 m/s

Height h=19m

Using second equation of motion

S=ut+0.5at²

Putting all values

19=23t-4.9t²

So this quadratic equation in T

So it's solutions

$t=\frac{4.69 ±\sqrt{21.97-15.52}}{2}\\=\frac{4.69 ±\sqrt{6.45}}{2}\\=\frac{4.69 ±2.54}{2}$

Taking positive sign

$t=\frac{4.69 +2.54}{2}=3.61sec.$

By negative sign

t=1.07sec