Question

A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)

a. The time at which the maximum height is achieved is functionsequation editor s.
b. The maximum height achieved by the projectile is functionsequation editor m.
c. The time when the projectile hits the ground is functionsequation editor s.
d. The range of the projectile is functionsequation editor m.
e. The speed of the projectile on impact with the ground is functionsequation editor m/s.

Answer 1

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Step-by-step explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

`t=(usin\alpha )/(g)`

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

`t=40sin30/9.81\\t=2.0secs`

b. the expression for the maximum height is expressed as

`H=(u^(2)sin^(2)\alpha )/(2g) \\H=(40^(2)0.25 )/(2*9.81) \\H=20.4m`

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

`R=(U^(2)sin2\alpha)/(g)\\R=(40^(2)sin60)/(9.81)\\R=141.2m`

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°