Question

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

Answer 1

Answer:

169.4 m/s

Explanation:

Given that the angle of projectile is θ_1 =450°

The speed of body at maximum height is U cosθ_1 = 150 m/s

The angle in second trail is θ_2 =37°  

From the given data U cosθ_1 = 150 m/s

U = 150 m/s / cosθ_1

=  150m/s / cos45°

=212.13 m/s

The velocity of the projectile at maximum height in second trail= Ucos(θ_2)

=212.13 m/s×cos37°

=169.4 m/s

Answer 2

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same