Projectile

[BobbyJones]

A projetcile is fired at an angle of 60 degrees, from the horizontal 1meter in height from the ground. The initial velocity is 40 m/s. The projetcile is aimed at the surface of a cliff of height 37m from the horizontal.


a) The maximum height reached by the projetcile with respect to the ground ?

My answer: 62.16 meters

b) The horizontal distance S travelled by the projetcile from the firing point to the point where the projetcile lands on top of the cliff?

My answer: ?, I got the heightest point time as 3.53 seconds, because at full range it would be t = 7.06 s. But the cliff is in the way, how do I work out the time from teh hightest point down to the landing point on the cliff?

 

[Deleted member 4993]

A projetcile is fired at an angle of 60 degrees, from the horizontal 1meter in height from the ground. The initial velocity is 40 m/s. The projetcile is aimed at the surface of a cliff of height 37m from the horizontal.


a) The maximum height reached by the projetcile with respect to the ground ?

My answer: 62.16 meters

b) The horizontal distance S travelled by the projetcile from the firing point to the point where the projetcile lands on top of the cliff?

My answer: ?, I got the heightest point time as 3.53 seconds, because at full range it would be t = 7.06 s. But the cliff is in the way, how do I work out the time from teh hightest point down to the landing point on the cliff?

What is the equation of the trajectory of the projectile?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[soroban]

Hello, BobbyJones!

A projectile is fired at an angle of 60 degrees, from the horizontal 1m in height from the ground.
The initial velocity is 40 m/s. The projectile is aimed at the topof a cliff of height 37m.

a) The maximum height reached by the projectile with respect to the ground ?

My answer: 62.16 meters

I assume you have the standard projectile functions:

. . \(\displaystyle x \:=\40\cos60^o)t \:=\:20t \)
. . \(\displaystyle y \:=\:1 + (40\sin60^o)t - 4.9t^2 \:=\:1 + (20\sqrt{3})t - 4.9t^2\)

Maximum height occurs when \(\displaystyle y' = 0.\)

. . \(\displaystyle 20\sqrt{3} - 9.8t \:=\:0 \quad\Rightarrow\quad t \:=\:\dfrac{20\sqrt{3}}{9.8} \:=\:3.534797566\:\approx\:3.535\text{ sec.}\)


Then: .\(\displaystyle y \:=\:1 + (20\sqrt{3})(3.535) - 4.9(3.535^2) \:=\:62.2244896 \)

Maximum height is about 62.22 meters.

b) The horizontal distance S travelled by the projectile from the firing point
to the point where the projectile lands on top of the cliff?

My answer: ?
I got the highest-point time as 3.53 seconds, because at full range it would be t = 7.06 s.

But the cliff is in the way, how do I work out the time from the hightest point
down to the landing point on the cliff? . Not necessary.

When is the height of the projectile equal to 37 meters?

Solve: .\(\displaystyle 1 + (20\sqrt{3})t - 4.9t^2 \:=\:37\)


You will get two answers.
(One is before it reaches maximum height.)

 

[BobbyJones]

So,

0 = -4.9t^2 + 34.63t-36

t= 1.29
and 5.76
Therefore

s = Uh x t
s = 20 x 5.76s = 115.2m

Have I got this right as the horizontal distance from the firing point 1m meter off the ground and landing on top of a cliff 37 meters high.