M2 SUVAT question

Hi, i am doing self study of M2 during the holidays and i am a bit stuck on these questions.

1) An aeorplane is climbing at an angle of 2° while maintaining a speed of 400ms-1. A package is released and travels a horizontal distance of 2500m before hitting the ground. by modelling the package as a particle projected with initial velocity the same as the velocity of the aeroplane, find the height of the aeroplane above the ground at the moment the package was released.
(i got the answer to be about 87.3m but the answer is 104m)

2) A child throws a ball with a speed of 20ms-1 from a window 5m above horizontal ground. If the ball hits the ground 1.5s later find the direction of projection.

3) A stone if projected from a point 0 on a cliff with the speed of 20ms-1 at an angle of elevation of 30°. T seconds later the angle of depression of the stone from 0 is 45°. Fidnt he value of T.

Thanks

Reply 1

Christophicus

1) An aeorplane is climbing at an angle of 2° while maintaining a speed of 400ms-1. A package is released and travels a horizontal distance of 2500m before hitting the ground. by modelling the package as a particle projected with initial velocity the same as the velocity of the aeroplane, find the height of the aeroplane above the ground at the moment the package was released.
(i got the answer to be about 87.3m but the answer is 104m)

Horizontal motion
s = 2500m
u = 400cos2 ms^-1
t = T
a = 0ms^-2

s = uT + ½aT²
s = uT + 0 = uT
T = s/u = 2500/400cos2

Vertical motion
s = S
u = 400sin2 ms^-1
a = -9.8 ms^-2
t = T

S = uT + ½aT²
S = 400sin2(2500/400cos2) + ½(-9.8)(2500/400cos2)²
S = -103.9 ≈ -104m
(-ve because the package falls below the zero point which is taken as the point at which the package is released)

Reply 2

Christophicus

2) A child throws a ball with a speed of 20ms-1 from a window 5m above horizontal ground. If the ball hits the ground 1.5s later find the direction of projection.

Vertical motion
u = 20sinθ ms^-1
s = -5m
a = -9.8ms^-2
t = 1.5s

s = ut + ½at²
-5 = 20sinθ(1.5) + ½(-9.8)(1.5)²
-5 = 30sinθ - 11.025
30sinθ = -5+11.025
30sinθ = 6.025
sinθ = 6.025/30
θ = arcsin(6.025/30) = 11.59° ≈ 11.6° above horizontal

Reply 3

Christophicus

3) A stone if projected from a point O on a cliff with the speed of 20ms-1 at an angle of elevation of 30°. T seconds later the angle of depression of the stone from O is 45°. Fidn the value of T.

The diagram shows that the angle of depression of the stone from O is 45°. Therefore the triangle that includes the side S_H and S_V is right angled and isosceles, so S_H = S_V.

Horizontal motion
s = S_H
u = 20cos30 ms^-1
a = 0ms^-2
v = u
t = T

s = ut
S_H = 20cos30(T) (1)

Vertical motion
s = -S_V = -S_H
u = 20sin30 ms^-1
a = -9.8ms^-2
t = T
v = ?

s = ut + ½at²
-S_H = 20sin30(T) + ½(-9.8)(T)² (2)

(1) = (2)
20cos30(T) = -(20sin30(T) - 4.9T²)
20cos30(T) = -20sin30(T) + 4.9T²
4.9T² - 20cos30(T) - 10T = 0
T(4.9T - 20cos30 - 10) = 0

T = 0 OR (4.9T - 20cos30 - 10) = 0

4.9T = 20cos30 + 10
T = (20cos30 + 10)/4.9
T = 5.58 seconds