Question

An object is projected from the ground with an initial velocity of 26.9 m/s at 50 degrees above the horizontal.
Find:
a. time of flight
b. range
c. maximum height

Answer 1

Given:

An object is projected from the ground with an initial velocity of 26.9 m/s at 50 degrees above the horizontal.

To Find:

What is value of Time of flight  ,Range and Maximum height?

Solution:

$\mathbf{Initial \ velocity= u = 26.9 \ \ \frac{m}{s}}$

$\mathbf{Angle\ with\ horizontal =\Theta =50^{\circ} }$

$\mathbf{Gravitational \ acceleration = g = 9.81 \ \ \dfrac{m}{s^{2}} }$

We know the formula of time of flight:

$\mathbf{Time\ of\ flight = T =\dfrac{2\times u\times \sin \Theta }{g} }$

$\mathbf{Time\ of\ flight = T =\dfrac{2\times 26.9\times \sin 50 }{9.81} }$

Time of flight = T = 4.2 second

We know the formula of range:

$\mathbf{Range= R =\dfrac{u^{2}\times \sin 2\Theta }{g} }$

$\mathbf{Range= R =\dfrac{26.9^{2}\times \sin (2\times 50)}{9.81} }$

Range of flight = R = 72.64 metre

We know the formula of maximum height:

$\mathbf{Maximum\ height= H_{max} =\dfrac{u^{2}\times \sin^{2} \Theta }{2\times g} }$

$\mathbf{Maximum\ height= H_{max} =\dfrac{26.9^{2}\times \sin^{2} 50 }{2\times 9.81} }$

$\mathbf{Maximum\ height= H_{max} =21.64 \ metre }$

Value of Time of flight  ,Range and Maximum height are 4.2 s , 72.64 m and 21.64 m.

Answer 2

An object is projected from the ground with an initial velocity of 26.9 m/s at 50 degrees above the horizontal.

To Find:

What is value of Time of flight ,Range and Maximum height?

Solution:

We know the formula of time of flight:

Time of flight = T = 4.2 second

We know the formula of range:

Range of flight = R = 72.64 metre

We know the formula of maximum height:

Value of Time of flight ,Range and Maximum height are 4.2 s , 72.64 m and 21.64 m.