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.

When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal.

At its highest point in this trajectory, the velocityof the projectile would be what?

Answer 1

169.4 m/s

Step-by-step 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

Step-by-step explanation:

Given

For first case

launch angle
`\theta =45^(\circ)C`

at highest point
`h=150 m/s`

`150=u\cos 45`

`u=(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* \cos 37`

`=169.41 m/s`

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