Physics Question Help (HSC) (1 Viewer)

[L C]

Hey, Can anyone help me out on how to answer this question. The answer is B, but i'm not sure how to get there...

2015 HSC QUESTION 20
A projectile was launched from the ground. It had a range of 70 metres and was in the air for 3.5 seconds. At what angle to the horizontal was it launched?

(A) 30°
(B) 40°
(C) 50°
(D) 60°

Help is much appreciated

 

[jazz519]

In these kind of questions you should draw a diagram of the starting information

Range = 70m. Can use that to find the initial horizontal velocity (using delta (x) = Ux t)

Ux = 70/3.5 = 20 m/s

Use the equation S = Uy t + 1/2 a t^2, to find Uy
Sub t = 3.5, a = -9.8 S = 0 (because this is the vertical displacement at the end the projectile hits the ground again so its back to the initial vertical displacement)

0 = Uy (3.5) + 1/2 (-9.8) (3.5)^2
Rearrange for Uy, so Uy = 4.9 x (3.5)^2 / (3.5) = 17.15 m/s

For the angle think of your sin, cos, tan rules

Here we know the opposite and adjacent so you use tan.

tan(theta) = Uy / Ux
tan(theta) = 17.15 / 20
theta = tan^-1 (17.15/20)
theta = 40.6 degrees

 

[dasfas]

u is initial V, a is the angle

x = u*t cos(a)

y = u*t sin(a) - 1/2*g*t^2

subbing in x = 70, y = 0, t = 3.5, g = 9.8

70 = u*3.5*cos(a)

0 = u*3.5*sin(a) - 1/2*9.8*(3.5)^2

60.025 = 3.5*u*sin(a)

u = 20/cos(a)

hence 60.025 = 3.5*20*tan(a)

solving for tan(a) gives 40.61 degrees

I did this using the 3u maths projectile motion rather than physics but both work

 

[L C]

Thank you so much @jazz519 and @dasfas!!