Question

Problem 9: A ball is thrown straight up at time t = 0 with an initial speed of 15 m/s. Take the point of release to be y0 = 0 and upwards to be the positive direction. No Attempt No Attempt 13% Part

(a) Calculate the displacement at the time of 0.50 s. No Attempt No Attempt 13% Part

(b) Calculate the velocity at the time of 0.50 s. No Attempt No Attempt 13% Part

(c) Calculate the displacement at the time of 1.0 s. No Attempt No Attempt 13% Part

(d) Calculate the velocity at the time of 1.0 s. No Attempt No Attempt 13% Part

(e) Calculate the displacement at the time of 1.5 s. No Attempt No Attempt 13% Part

(f) Calculate the velocity at the time of 1.5 s. No Attempt No Attempt 13% Part

(g) Calculate the displacement at the time of 2.0 s. show answer No Attempt 13% Part

(h) Calculate the velocity at the time of 2.0 s.

Answer 1

Date: 11 Page No. Ans. given u=15ms throum upwards, so g=-9.8 m/s2 Can t = 0. 505 S = 1 g ² + ut S = (15)(0:59x1 698960592 S = 7.5 -T-225 S = 6.275 m. • ball is thrown straight upwards, so travelled distance = displacement distalacement = 6.275 m 06) Velocity at t=0.505 = ut gt v=15+(-9-890.5) v=15-4.9 Iv=11.1 Velocity at t=0.5 Setesen is 101 m/s. CC) displacement at t=10S S = 1 g + ² + ut S = 150) + (-9-890 - 3 = 15-0,3 = |-| Ý

Date: 1 Page No. distance travelled at t=1.08 = displacement displacement = 11.1 m D) Velocity at t- tiosec V = ut gt V = 15+(-9.8) (1) [V = 5.2 m/s LED displacement at t=1.5 sec S = ut+ gt? S= (15)(1-5)+469.89(1:5) S = 22.5+5 (-9,8)(225) S = 22.5-11025 S = 11.475 m velocity at t= 1.5 see = ut gt v = 15-9:8X1.5 V = 15-14.7 Tv=0.3 m/

Date: / Page No. displacement at t=2 see on o. We know that v=0 m/s, u=15 m/s, g = 9.8m. its time taken by ball to reach at maximum height will be V= utgt O = 15+(-9.8 xt t = 15 sec 9.8 t=1.53 sec So at t=2 sec, the ball will be back towards the earth - 2-1.53 = 0.47 sec So now, the displacement, will not be egual to the distance S=ut I gt 2 0 0 4 0 m/s in this case S = 1 / 2 g 4² g is posit because com S = 1 89.84 (0.1472 / because com downward 3= 1.082 m

Date: 11 Page No. Velocity at t=asec. U=0,9 = 98 mis? v=utgt v=o+(9.8 (0.47) 1 = 4.606 m/s So, at t= 28ee relouty will be 4. 606 ms.