A projectile is fired at 45 degree with the horizontal with a velocity of 20 m/s

Question

A projectile is fired at 45 degree with the horizontal with a velocity of 20 m/s , draw the net velocity vectors at points A, B and the highest point C (b) net velocity magnitude at A and B is same or different, qualitative answer no numbers (c) give the net velocity magnitude at C. A ball is dropped from a height H when it hits the ground it bounce upwards with half the velocity with which it hit the ground and attains a height of 5.0 m , what was the height it was dropped from. In this problem use g = 10.0 m/s2. An electron traveling with a velocity of 2.1 * 109 cm/s at point A it is given an upwards acceleration of 5.3 * 1017 cm/s2 (a) how long it takes for electron 6.2 cm from point A (b) what vertical distance it covers in this time (c) what is the displacement of electron in this time.

Explanation / Answer

the velocity vector at the given points will be along the tengent to the curve of projectile

b)

the hight of the points A & B are same so by the consrvation of the energy

E total = P.E+ K.E.

since the hight is same the P.E. =mgh is same for A& B

so

(P.E)_A + (K.E)_A = (P.E.)_B + (K.E.)_B

so

(K.E)_A = (K.E)_B

since K.E = mv^2/2

since mass is constant so the magnitude of the velocities at A&B will be constant but the direction will be along the tangent.

(C) at max hight the virtical velocity of the particle is =0 and the horizontal velocity remains constant throughout the flight so v_c = 20cos45 = 20sqrt(2) =28.28m/s

2)use eq of motion

in bouncing back

(v^2 = u^2 -2gS) v=0 at max hight

(u=sqrt{2 imes 10 imes 5} = 10m/s)

so the velocity before bouncing was 20m/s

using same eq since u=0 at time of starting

(S= v^2/2g = 400/20 = 20 m) since the velocity was downwards -iv sign get canceled.

3) a) since the a is vertical so v_horizontal is cons

s=vt and t=S/v=6.2/(2.1x10^(9)) = 2.95x10^(-9)s

b) (S = ut + at^2/2) since u_vertical = 0

S_Vert= 5.3 *10^17 *(2.95*10^(-9))^2/2= 2.3cm

c)

total displacement = (sqrt{x^2 +y^2} = sqrt{6.2^2 + 2.3^2} = 6.612 cm)

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